Assignment To Expression With Array Type

Arrays in C: Pitfalls and Solutions

Understanding Array Decay

In the C programming language, arrays are treated as pointers to their first element. This means that when an array is passed to a function, the function receives a pointer to the array, not a copy of the array itself. This behavior, known as array decay, can lead to confusion when working with arrays.

For example, the following code attempts to assign the value of one array to another array:

int a[10] = {1, 2, 3, 4, 5}; int b[10]; b = a;

This code will result in a compiler error because arrays are not assignable in C. Instead, to copy the values of one array to another, you can use the memcpy function:

memcpy(b, a, sizeof(a));

Note that memcpy works by copying the entire array, including its elements, whereas the assignment operator = only copies the pointer to the first element of the array.

Using Strings

When working with strings, you can use the strcpy function to copy the value of one string to another. The strcpy function takes two arguments: a destination string and a source string, and it copies the contents of the source string to the destination string:

char str1[100] = "Hello"; char str2[100]; strcpy(str2, str1);

Error: Assignment on Expression with Array Type

Another common error that can occur when working with arrays is the Assignment on Expression with Array Type error. This error occurs when you attempt to assign a value to an entire array, as in the following code:

int a[10]; a = {1, 2, 3, 4, 5};

To fix this error, you need to use the = operator to assign each element of the array individually:

int a[10]; a[0] = 1; a[1] = 2; a[2] = 3; a[3] = 4; a[4] = 5;