error invalid array assignment c

C++ 数组分配错误 : invalid array assignment

标签 c++ c arrays

我不是 C++ 程序员，所以我需要一些有关数组的帮助。 我需要将一个字符数组分配给某个结构，例如

我得到 error: invalid array assignment

如果 mStr.message 和 hello 具有相同的数据类型，为什么它不起作用？

因为您不能分配给数组——它们不是可修改的左值。使用 strcpy:

正如 Kedar 已经指出的那样，您还写出了数组的末尾。

关于C++ 数组分配错误 : invalid array assignment，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/4118732/

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Programmer guide, tips and tutorial, c++ bug: [error] invalid array assignment.

C++BUG: [Error] invalid array assignment

1. Introduction2. The difference between the return value of memcpy() function prototype function header file and strcpy

1. Introduction

When using array to assign value to array, the above bug will appear. The general chestnut is as follows:

The purpose is to store the data in the structure array object SS [J], but this assignment will report an error.

2. memcpy()

Function prototype

The data of consecutive n bytes with SRC pointing address as the starting address is copied into the space with destination pointing address as the starting address.

Header file

The use of # include & lt; string. H & gt;;

Both # include & lt; CString & gt; and # include & lt; string. H & gt; can be used in C + +

Return value

Function returns a pointer to dest.

The difference from strcpy

1. Compared with strcpy, memcpy does not end when it encounters’ \ 0 ‘, but will copy n bytes. Therefore, we need to pay special attention to memory overflow. 2. Memcpy is used to copy memory. You can use it to copy objects of any data type, and you can specify the length of the data to be copied. Note that source and destination are not necessarily arrays, and any read-write space can be used; however, strcpy can only copy strings, and it ends copying when it encounters’ \ 0 ‘.

For 2D arrays

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Problem with arrays

I'n trying to make a sketch about a led matrix 8x8 with 2 74hc595 shift register, but i have a problem with one array. I have several arrays declared like this:

int M[8][8]= { {0, 1, 1, 1, 0, 1, 1, 1}, {0, 0, 1, 0, 0, 1, 1, 1}, {0, 1, 0, 1, 0, 1, 1, 1}, {0, 1, 0, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1}, {0, 1, 1, 1, 0, 1, 1, 1} };

and i want to make other array like this: const int LETRAS=25; int frase[LETRAS][8][8]={W,W,WW,P,R,A,C,T,I,C,A,N,D,O,A,R,D,U,I,N,OO,C,O,M,SPACE};

but i have this error 200: error: invalid conversion from 'int (*)[8]' to 'int'

If i try this way

int frase[LETRAS][8][8]; frase[0]= M;

i have the follow error: 203: error: invalid array assignment

Does anybody can help me please?

First things first - if all you saving in each element is a 1 or a 0, that's a huge waste of memory. Make the elements "byte" or better still, pack the bits.

And post your code - use code tags.

You've got two dimensions to that array, but you're only specifying one. That may "work" but it will bite you later.

Your biggest problem is that you are trying to shove a letter in a hole meant for a number. You have an int data type and then try to put M in it.

Maybe you want the ascii representation of M, so you could use 'M'. I'm not real sure what you want to do so I'm not sure how you should change it. But what you've got now is just completely wrong on too many levels to list.

Re-reading it, it isn't just a waste of memory, unless you're running it on a Mega, it's almost certainly more memory than you've got. So, "byte" rather than "int", and then eventually, you'll be moving the whole lot into PROGMEM.

i want to store in an array 25 matrix 8x8, each matrix position is an integer, so i will have 25 matrix like M (with different digits), so my question is how can i make this?

i want to store in an array 25 matrix 8x8, each matrix position is an integer,

You probably can't. That would be 3200 bytes, which is more than exists on an Arduino.

If you want to have an array of references to bitmaps, suitable for displaying on dot-matrix displays, that usually comes out somewhat differently. Keeping with the basics you have now, You could do:

More typically, one would pack bits into a sort of font table, stick that in program memory, and index it more-or-less directly from character values:

undesrtood, thanks to all, i'll try this way

Related Topics

Array type char[] is not assignable – C

Quick Fix: Utilize strcpy to copy a string to a character array. Example: strcpy(word, "Jump"); .

The Problem:

You have a C program that prompts the user to choose from three options: "Jump", "Run", or "Dance." Based on the user’s choice, you want to assign the corresponding string to the array word . However, you are encountering the error "array type char[30] is not assignable" when trying to make this assignment.

The Solutions:

Solution: 1: use `strcpy` function to copy string to character array.

The error “array type `char[30]` is not assignable” occurs because you cannot directly assign a string literal to a character array in C. To assign a string literal to a character array, you need to use the `strcpy` function. The `strcpy` function takes two arguments: the destination character array and the source string literal. It copies the contents of the string literal to the character array.

Here’s the corrected code:

With this correction, the program will successfully assign the appropriate string to the word character array based on the user’s choice, and then print out the corresponding action.

Solution 2: The Concept of Array Name And Assignment Operators

An array name is not a modifiable lvalue, according to the C11 standard. As a result, you can’t use the assignment operator (=) on it. To copy the data into the array, you must use strcpy() from string.h (char array) or memcpy() in general. “`C++ #include #include //Including the string.h header for strcpy usage int main() { int choice1; char word[30]; //An array named word is declared to store user’s choice printf(“You have three choice.\n”); printf(“[1] Jump [2] Run [3] Dance\n”); scanf(“%d”, &choice1); //Reads user input and stores it in choice1

Solution 3: Pointer cannot be assigned to array object

The key point to understand here is that an array expression in C is treated as a non-modifiable lvalue, meaning it refers to an object in memory but cannot be assigned to. The given code attempts to assign a string literal, which is an array of characters, directly to an array variable, which is not allowed in C.

In C, the assignment operator (=) can only be used to assign values to individual array elements or to copy the contents of one array to another using library functions like `strcpy` or `memcpy`. This is because arrays in C are not pointers, but rather contiguous blocks of memory that store elements of the same type. An array expression, such as `word`, is essentially a pointer to the first element of the array, and it cannot be assigned to another array object.

To resolve the errors in the given code, you need to use one of these library functions to copy the string literals to the array `word`. Here’s a corrected version of the code:

In this corrected code, the `strcpy` function is used to copy the string literals to the `word` array. The `strcpy` function takes two arguments: the destination array (where the string literal will be copied) and the source string (the string literal itself). When the `strcpy` function is called, it copies the contents of the source string to the destination array, including the null-terminator character (‘\0’) that marks the end of the string.

Solution 4: Use strcpy from

The issue here is that you’re trying to directly assign a string literal to a character array, which is not allowed in C. Instead, you need to use the `strcpy` function from the `<string.h>` header to copy the contents of the string literal into the character array.

Here’s how you can fix the code using `strcpy`:

With this change, the code should work as expected.

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C/C++BUG: [Error] invalid array assignment

C++ array assign error: invalid array assignment

Because you can’t assign to arrays — they’re not modifiable l-values. Use strcpy:

And you’re also writing off the end of your array, as Kedar already pointed out.

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