assignment of read only location c vector

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Home » C programs » C common errors programs

Error: Assignment of read-only location in C

Here, we are going to learn why an Error: Assignment of read-only location in C occurs and how to fixed in C programming language? By IncludeHelp Last updated : March 10, 2024

• Error: Assignment of read-only variable in C

Error: assignment of read-only location occurs when we try to update/modify the value of a constant, because the value of a constant cannot be changed during the program execution, so we should take care about the constants. They are read-only.

Consider this example:

Here, the statement str[0]='p' will try to change the first character of the string, thus, the "Error: assignment of read-only location" will occur.

How to fix?

Do not change the value of a constant during program execution.

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assignment of read-only location (vectors)

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Hey guys, I just finished semester one at school, and I have been doing some coding exercises to keep sharp over the winter break, and I have been some new errors on this exercise. This exercise came from a "top-coder" practice problem that was used a few years ago. Here is the problem statement: Code: ........ A chessboard pattern is a pattern that satisfies the following conditions: The pattern has a rectangular shape. The pattern contains only the characters '.' (a dot) and 'X' (an uppercase letter X). No two symbols that are horizontally or vertically adjacent are the same. The symbol in the lower left corner of the pattern is '.' (a dot). You are given two ints rows and columns. Write a method that computes the chessboard pattern with these dimensions, and returns it in a vector <string>. The elements of the return value correspond to rows of the pattern. Specifically, the first character of the last element of the return value represents the lower left corner I addressed the problem by writing this function here: Code: vector <string> makeChessboard (int w, int h){ string row = ""; vector<string>Board(); int i = 0; for(int height = 0; height < h; height++){ while(i<(w*h)){ for(int j = 0; j<w; j++){ if (i % 2 == 0){ row[j] = '.'; } else{ row[j] = 'X'; } i++; } } Board[height] = row; } return Board; } I am getting many errors on line 23 were I say Code: Board[height] = row; , including: Code: |23|warning: pointer to a function used in arithmetic |23|error: assignment of read-only location '*(Board + ((unsigned int)height))'| |23|error: cannot convert 'std::string' to 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >()' in assignment| |25|error: conversion from 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > > (*)()' to non-scalar type 'std::vector<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::allocator<std::basic_string<char, std::char_traits<char>, std::allocator<char> > > >' requested| I have no clue why I am getting those errors, I have never ran into the first warning before, could someone please shed some light on what I am doing wrong? I tried using push back also to no avail. I don't know why it is complaining about trying to convert between types. If you guys could help out with this I would appreciate it a ton! Thanks a ton guys! I appreciate it!

Last edited by jlangfo5; 12-16-2010 at 09:30 PM .

Code: vector<string>Board(); Here you declare Board to be a function returning a vector<string> (much like the function you're in, as a matter of fact), not an object of that type. If you want an empty constructor list, then simply leave off the parentheses.

Well thanks tabstop! I was able to fix that and it then compiled, but then everyone's favorite, the segfault appeared. I tried to test my function to see if it would work correctly, with this code here. Code: #include <iostream> #include <vector> using namespace std; vector <string> makeChessboard (int w, int h){ string row = ""; vector<string>Board; int i = 0; for(int height = 0; height < h; height++){ while(i<(w*h)){ for(int j = 0; j<w; j++){ if (i % 2 == 0){ row[j] = '.'; } else{ row[j] = 'X'; } i++; } } Board[height] = row; } return Board; } int main() { int h = 8; int w = 8; vector<string>chess; cout<< "making chess"; chess = makeChessboard (h, w); cout << "chess made"; for(int i = 0; i<chess.size(); i++){ cout << chess[i] << endl;} return 0; } It outputs "making chess", but crashes when it tries to do the assignment. So, it never gets to "chess made". Any takers? Thanks once again guys!

You are using the vector without first resizing it to appropriate size. Also, you forgot to include <string>. Same thing with the string. You are trying to access individual characters when it's empty. Logic is also wrong. You seem to be doing it in a overcomplicated way. I'd sit down and think a little more about how to accomplish it.

Last edited by Elysia; 12-17-2010 at 03:55 AM .

Originally Posted by Adak io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions. Originally Posted by Salem You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much. Outside of your DOS world, your header file is meaningless.

Thanks Elysia! It compiled after I used push back on the string instead of trying to assign to it like i was! Now I just have to fix a logic that error that isn't so bad, and it will be working properly! I'll post it when i have it working right!

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Pointers in C / C++ [Full Course] – FREE

Const Pointer in C++

In this lecture, we discuss about Const pointer and Pointer to Const and Const Pointer to Const in C++.

Why do we need Const Pointer or Pointer to Const?

A pointer is a variable that stores the memory address of another variable or the memory address of the memory allocated on the heap. For example,

Here, we have created a variable ‘num’ and initialized it with the value 11, then created a pointer ‘ptr’ and initialized it with the address of variable ‘num’. Now, the pointer ‘ptr’ contains the memory location where the value of the variable ‘num’, i.e. 11, is stored.

You can access this value using the pointer ‘ptr’ by dereferencing it. Like this,

You can also change the value at that memory location. For example, you could change the value at the memory location pointed to by ‘ptr’ to 20, which in turn would change the value of variable ‘num’. You can print and confirm that the value of the variable ‘num’ is now 20.

This is how you change the value at a memory location using a pointer.

Now, suppose you pass a pointer to a function and, inside the function, someone changes the value. This value change inside the function using the pointer will be reflected outside the function as well. For example, if you have a variable ‘num’, initialized with the value 11, and then create a pointer ‘ptr’ containing the address of this variable, if you then pass this pointer to the function ‘getSquare’, this function is supposed to return the square of the value at the memory location pointed to by the given pointer. But if a new developer comes along and tries to change the value at this location pointed to by the pointer, because you passed a pointer, he can directly change the value.

We wanted the square of 11, but inside the function value at memory location got changed to 2, and it returned the square of 2 instead. Also after the function getSquare() returns, the value of num is also changed to 2.

Pointer to Const in C++

You can restrict the user from changing the value pointed to by the pointer by using a pointer to const. While creating a pointer, you can make the pointer point to a const integer. In both cases, the pointer ‘ptr’ will be pointing to a memory location, but it cannot change the data at that memory location. So, if someone tries to change the value now, it will give an error. Similarly, if you want to pass a pointer to a function for read operation only, then you can pass it like this.

Here, you are passing a pointer pointing to a const integer, so nobody can change the value at that memory location using the pointer. This is called pointer to const.

Even if you restricted the user so that they cannot change the value at the memory location pointed to by the pointer ‘ptr’, the pointer itself is also a variable, so you can change it to point to a new memory location. Check out this example: here, inside the ‘getSquare()’ function, we change the value of the pointer ‘ptr’ and make it point to a new memory location.

Is there any way to restrict the user so that they can’t change the pointer to point to a new memory location? Answer is Yes, by making pointer a constant pointer.

To restrict the user so they cannot make the pointer point to another memory location, you need to make the pointer const, like this.

Here, your pointer is a const pointer pointing to a memory location, or the address of variable ‘num’. Now, even if you want to point the pointer to a new memory location, you can’t do that because the pointer is of const type. However, using the pointer, you can change the value at that memory location, as the pointer is not pointing to a constant integer. This is different from the previous scenario.

Const pointer pointing to Const in C++

In this section, we will cover a combination of both: a const pointer pointing to a constant integer. Suppose you initialize the variable ‘num’ with the value 11. Now, you want to create a pointer to this but you want to restrict it so that no one can change the value of ‘num’ using the pointer, and also, once you initialize the pointer to the address of ‘num’, no one can make that pointer point to any other location. For that, you have to use a const pointer pointing to a const value, like this.

Here, your pointer ‘ptr’ is a const pointer and is pointing to a constant value. So, we cannot make the pointer point to any other location, and we also cannot use the pointer to change the value at that memory location.

Const pointer vs Pointer to Const vs Const Pointer to Const

Sure, here’s a simple table to illustrate the differences:

• A const pointer means that the memory location that the pointer is pointing to can be changed, but the pointer itself cannot change and point to a different memory location. In other words, the pointer is constant, but the data at which it points is not.
• A pointer to const means that the pointer can point to different locations, but the data at the memory location the pointer is pointing to cannot be changed using this pointer. In other words, the data is constant but the pointer is not.
• A const pointer to const combines both concepts – the pointer cannot point to a different memory location and the data at the memory location cannot be changed. Both the pointer and the data it points to are constant.

In this lecture, we learned about Const pointer and Pointer to Const and Const Pointer to Const in C++.

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On compiling the program below the error "assignment of read-only location '1107307520u->COUNT16.INTFLAG.bit.OVF' " appears

The TC_INTFLAG register is not write protected

Any good ideas?

The processor is a SAMD21G18A. THe compile error does not appear when using a similar statement with another TC3 register

 #include "sam.h "

 void TC3_Handler(void) {   if (TC3->COUNT16.INTFLAG.bit.OVF) // TC3 overflow interrupt?   { TC3->COUNT16.INTFLAG.bit.OVF = 1u;    /* Clear the interrupt flag. This line causes the error */      TC3->COUNT16.INTFLAG.reg |= TC_INTFLAG_OVF; // No compile error!   } }

int main(void) {     /* Initialize the SAM system */     SystemInit();

    /* Replace with your application code */     while (1)      {     } }

The TC_INTFLAG_Type typedef in tc.h has this comment

So this is about avoiding a read-modify-write (that is the likely implementation of bit assignment) that would cause all flags that are currently 1 to be cleared (instead of just clearing OVF in your case).

A read-modify-write (and thus clearing all set flags) is what you have in the compiling statement:

better use:

But obviously that only makes a difference if you are actually using other flags in the register.

Thanks again for the very useful and illuminating answer - I see that I'm still on the SAMD learning curve, the AT(x)mega MCUs clear the flag when they jump to the ISR

At the moment I'm about 400km away from the hardware, so I'll have to wait for a test.

On p. 670 of the SAMD21 datasheet it says that writing a 0 to an INTFLAG bit has no effect, so if other bits are set your suggested code:

will leave them set.

C++ – “error: assignment of read-only location” in unordered_map (C++)

c iterator unordered-map

I have an awkward hash table (specifically, an unordered_map) with int keys and vector< vector< int >> data. I periodically need to update elements in this two-dimensional vector of ints. There's no intrinsic reason I shouldn't be able to, right? A newer g++ compiler I've switched to complains of an assignment of read-only location on the line designated below.

I'm new to C++, so nothing's too obvious. Thank you for any help.

Following Tim's suggestion

I've replaced the relevant parts of code above with the following:

This code compiles without the error and appears to run fine. Like Tim, I still don't quite understand why the fix works. The error was previously appearing with gcc version 4.1.2 20080704 (Red Hat 4.1.2-44) but not with gcc version 4.0.1 (Apple Inc. build 5465). I will try to dissect the error more carefully when I'm not under a tight deadline!

Best Answer

Are you sure there really are thisStep + 1 elements in every first-level vector and NUM_DEMES elements in every second-level vector?

You're not actually assigning to a map iterator, if I read correctly, so I suspect the error is in the vector access.

It may be helpful to break that last statement up into multiple statements so that only one thing is being done on each to narrow down where the problem is. e.g.,

By the way, piItr = phenotypeIs.begin(); has no effect here, it could be simply:

Related Solutions

C++ – why can templates only be implemented in the header file.

Caveat: It is not necessary to put the implementation in the header file, see the alternative solution at the end of this answer.

Anyway, the reason your code is failing is that, when instantiating a template, the compiler creates a new class with the given template argument. For example:

When reading this line, the compiler will create a new class (let's call it FooInt ), which is equivalent to the following:

Consequently, the compiler needs to have access to the implementation of the methods, to instantiate them with the template argument (in this case int ). If these implementations were not in the header, they wouldn't be accessible, and therefore the compiler wouldn't be able to instantiate the template.

A common solution to this is to write the template declaration in a header file, then implement the class in an implementation file (for example .tpp), and include this implementation file at the end of the header.

This way, implementation is still separated from declaration, but is accessible to the compiler.

Alternative solution

Another solution is to keep the implementation separated, and explicitly instantiate all the template instances you'll need:

If my explanation isn't clear enough, you can have a look at the C++ Super-FAQ on this subject .

C++ – ny advantage of using map over unordered_map in case of trivial keys

Don't forget that map keeps its elements ordered. If you can't give that up, obviously you can't use unordered_map .

Something else to keep in mind is that unordered_map generally uses more memory. map just has a few house-keeping pointers, and memory for each object. Contrarily, unordered_map has a big array (these can get quite big in some implementations), and then additional memory for each object. If you need to be memory-aware, map should prove better, because it lacks the large array.

So, if you need pure lookup-retrieval, I'd say unordered_map is the way to go. But there are always trade-offs, and if you can't afford them, then you can't use it.

Just from personal experience, I found an enormous improvement in performance (measured, of course) when using unordered_map instead of map in a main entity look-up table.

On the other hand, I found it was much slower at repeatedly inserting and removing elements. It's great for a relatively static collection of elements, but if you're doing tons of insertions and deletions the hashing + bucketing seems to add up. (Note, this was over many iterations.)

Related Topic

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cppreference.com

Std::vector<t,allocator>:: operator=.

Replaces the contents of the container.

[ edit ] Parameters

[ edit ] return value, [ edit ] complexity, [ edit ] exceptions, [ edit ] notes.

After container move assignment (overload (2) ), unless element-wise move assignment is forced by incompatible allocators, references, pointers, and iterators (other than the end iterator) to other remain valid, but refer to elements that are now in * this . The current standard makes this guarantee via the blanket statement in [container.reqmts]/67 , and a more direct guarantee is under consideration via LWG issue 2321 .

[ edit ] Example

The following code uses operator = to assign one std::vector to another:

[ edit ] See also

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Posted on Oct 28, 2020 • Updated on Sep 24, 2021

Let's try C++20 | std::span

Eventually, I have decided to try C++20 😀

Let's start with std::span !

What is a Span?

Here is the definition given by cppreference :

The class template span describes an object that can refer to a contiguous sequence of objects with the first element of the sequence at position zero.

Spans are sometimes called "views" because they don't own the sequence of objects.

If you are familiar std::string_view from C++17, then you can easily understand the concept of spans. std::string_view s are somehow spans on sequences of characters. The main difference is that spans can modify the objects while string views can't modify the caracters.

In the following, I will try to use std::span s in various situations to understand how they work and how they can be useful for future code.

With Plain Arrays

I personally use a lot of plain arrays because I work with on embedded systems where those C-style arrays are widely used. Why? Because dynamic allocation is forbidden (so std::vector s are not an option) and because there is a lot of C code.

When you pass a plain array to a function, it is decayed to a pointer to its first element and its size is lost. std::span s sound like a great solution to safely pass around plan arrays to functions.

First, let's write a template function that prints any span:

Then, let's create a span from a plain array:

This code outputs:

The size of the array is deduced from its type and stored in the std::span object.

With a Pointer and a Size

If you are already in a function where the original plain array has been "transformed" to pointer + size, you can create a span from them. Of course, you cannot be sure it maps the original plain array:

The last span is reading outside of the array, which is clearly an undefined behavior. You are responsible for creating the span with a valid sequence.

The form (2) of std::span 's constructors is used here:

With std::array s or std::vector s

You can also easily create spans from std::array s or std::vector s:

Obviously, constructor (5) is used for the array:

But which one is used for the vector?

It took me a while to understand that form (7) is used:

This is because a vector is a range :

std::ranges::range is a concept, a new feature of C++20. Concepts are one of the next features I will try.

With a part of an std::array or std::vector

You can't directly create a span over a part of the array or the vector. For instance:

doesn't compile because of the following error:

error: no viable constructor or deduction guide for deduction of template arguments of 'span'

The same error occurs with an std::vector .

The trick is to get an iterator/pointer from the container. For instance:

Here, constructors (2) and (3) are used:

It is possible to create subspans, which are spans on a part of a span. There are 3 member functions to do that:

• first() : obtains a subspan consisting of the first N elements of the sequence
• last() : obtains a subspan consisting of the last N elements of the sequence
• subspan() : obtains a subspan

Modify Objects in the Span

It is possible to modify the objects the span refers to:

Note than std::span doesn't have cbegin() and cend() as iterator functions. It only has begin() / end() / rbegin() / rend() .

Nevertheless, if we declare data as const std::array data = {1,2,3,4,5}; , then the code won't compile. Here is an example where we try to modify a const array using a span:

This doesn't compile:

error: assignment of read-only location 's.std::span<const int, 5>::operator[](0)'

Static vs Dynamic Extent

The size of the sequence of objects can be static or dynamic, as stated in the overview of the class :

A span can either have a static extent, in which case the number of elements in the sequence is known at compile-time and encoded in the type, or a dynamic extent. If a span has dynamic extent a typical implementation holds two members: a pointer to T and a size. A span with static extent may have only one member: a pointer to T.

Let's create a function to print the extent of spans:

Those 0xffffffffffffffff do look like std::dynamic_extent .

A dynamic extent really means that the size of the span changes along with the size of the sequence. Here is an example where we push items to a vector and look how the size of the span evolves (we know from the previous example it has dynamic extent):

Non-Template Code with Dynamic Extent

A span with a static extent can be converted implicitly to span with a dynamic extent:

Why is this really interesting? Because we can rely on this conversion to write non-template functions that take spans of any size as their parameters. This is something we cannot do with std::array s. Compare these 2 functions that prints arrays:

Both of course output "hello". Because the first one is non-template, it can be simply declared in a *.h and defined in a *.c. This is not possible of course for the second one. But I think you already know that "issue" with templates, right? 😅

And finally, because I am an embedded software developer and I care about footprint, let's take a look at the size of a span:

I didn't test the ROM footprint because I don't have a C++20 compatible embedded compiler on my computer yet.

OK, spans look amazing 😍

I believe they will help write me better code, mainly when I deal with plain arrays I get from C code. Spans will provide a good solution (safe, standard, portable) to carry the size of the data along with a pointer to them. They will probably be a nice bridge between the C and the C++ words.

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The type parameter <T> specifies the type of the vector. It can be any primitive data type such as int , char , float , etc. For example,

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Notice that we have not specified the size of the vector during the declaration. This is because the size of a vector can grow dynamically, so it is not necessary to define it.

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Here, we are initializing the vector by providing values directly to the vector. Now, both vector1 and vector2 are initialized with values 1 , 2 , 3 , 4 , 5 .

Here, 5 is the size of the vector and 12 is the value.

This code creates an int vector with size 5 and initializes the vector with the value of 12 . So, the vector is equivalent to

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Here, we have declared and initialized three different vectors using three different initialization methods and displayed their contents.

Basic Vector Operations

The vector class provides various methods to perform different operations on vectors. We will look at some commonly used vector operations in this tutorial:

• Add elements
• Access elements
• Change elements
• Remove elements

1. Add Elements to a Vector

To add a single element into a vector, we use the push_back() function. It inserts an element into the end of the vector. For example,

Here, we have initialized an int vector num with the elements {1, 2, 3, 4, 5} . Notice the statements

Here, the push_back() function adds elements 6 and 7 to the vector.

Note : We can also use the insert() and emplace() functions to add elements to a vector.

2. Access Elements of a Vector

In C++, we use the index number to access the vector elements. Here, we use the at() function to access the element from the specified index. For example,

• num.at(0) - access element at index 0
• num.at(2) - access element at index 2
• num.at(4) - access element at index 4

Note: Like an array, we can also use the square brackets [] to access vector elements. For example,

However, the at() function is preferred over [] because at() throws an exception whenever the vector is out of bound, while [] gives a garbage value.

3. Change Vector Element

We can change an element of the vector using the same at() function. For example,

In the above example, notice the statements,

Here, we have assigned new values to indexes 1 and 4 . So the value at index 1 is changed to 9 and the value at index 4 is changed to 7 .

4. Delete Elements from C++ Vectors

To delete a single element from a vector, we use the pop_back() function. For example,

In the above example, notice the statement,

Here, we have removed the last element ( 7 ) from the vector.

• C++ Vector Functions

In C++, the vector header file provides various functions that can be used to perform different operations on a vector.

• C++ Vector Iterators

Vector iterators are used to point to the memory address of a vector element. In some ways, they act like pointers in C++.

We can create vector iterators with the syntax

For example, if we have 2 vectors of int and double types, then we will need 2 different iterators corresponding to their types:

Initialize Vector Iterators

We can initialize vector iterators using the begin() and end() functions.

1. begin() function

The begin() function returns an iterator that points to the first element of the vector. For example,

2. end() function

The end() function points to the theoretical element that comes after the final element of the vector. For example,

Here, due to the nature of the end() function, we have used the code num.end() - 1 to point to the last element of the num vector i.e. num[2] .

Example: C++ Vector Iterators

In this program, we have declared an int vector iterator iter to use it with the vector num .

Then, we initialized the iterator to the first element of the vector using the begin() function.

Then, we printed the vector element by dereferencing the iterator:

Then, we printed the 3rd element of the vector by changing the value of iter to num.begin() + 2 .

Finally, we printed the last element of the vector using the end() function.

• Example: Iterate Through Vector Using Iterators

Here, we have used a for loop to initialize and iterate the iterator iter from the beginning of the vector to the end of the vector using the begin() and end() functions.

Also Read :

Table of Contents

• Introduction
• Example: C++ Vector
• Add Elements to a Vector
• Access Elements of a Vector
• Change Vector Element
• Delete Elements from C++ Vectors

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Build fails on PPC: simd_vector_scalar.h: error: assignment of read-only location '* data' #166

barracuda156 commented Jan 7, 2023

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barracuda156 commented Jun 5, 2023

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• Windows Programming
• UNIX/Linux Programming
• General C++ Programming
• Pointer to a function used in arithmetic

    Pointer to a function used in arithmetic

【C言語】assignment of read-only location ‘xxx’

“assignment of read-only location ‘xxx'” というエラーメッセージは、読み取り専用の場所に代入しようとした場合に表示されます。

例えば、次のようなコードを書いたとします。

このコードでは、定数 x の値を 10 から 20 に変更しようとしています。しかし、定数は値を変更することができないため、このコードは “assignment of read-only location ‘x'” というエラーを引き起こします。

正しいコードは、次のようになります。

この場合、定数 x の値を変数 y にコピーし、変数 y の値を変更しています。そのため、このコードはエラーを引き起こしません。