How to set value of all structure members at once?

Question

How can I set structure members using a single line of code?

typedef struct {
    int x;
    int y;
}test;

test obj[2];
int main()
{

 obj[0].x=0; //this works
 obj[0].y=0; //this works
 obj[0]={0,0};  //compilation error

}

Answer 1

Structs can be assigned to with the = operator, exactly the same way as any other value in C:

obj[0] = existing_test;
obj[1] = function_returning_test();

The above rely on struct values that have themselves come from somewhere else in the program (possibly having been initialized in multiple statements like in the question); to create a struct value in a single expression, use the object literal syntax:

obj[0] = (test){ .x = 15, .y = 17 };
obj[1] = (test){ .y = 19 };

Any fields left out of a such a literal are still present, but set to the appropriate zero value for their type, so in the above example obj[1].x is set to zero.

Answer 2

If you are looking to set the values to zero, the following line of code should work:

memset(&obj,0,sizeof(obj));

However, it is only going to work if you want the values initialized to zero

Answer 3

You can only initialize all values in your array of structs in a single line of code when the instance is created. (not including use of memset). Consider the following example:

#include <stdio.h>

typedef struct {
    int x;
    int y;
} test;

int main()
{
    test obj[2] = { { 3, 5 }, { 4, 6 } };
    int i = 0;

    for (i = 0; i < 2; i++)
        printf ("\n  obj[%d].x = %d  obj[%d].y = %d\n", i, obj[i].x, i, obj[i].y);

    return 0;
}

output:

obj[0].x = 3  obj[0].y = 5

obj[1].x = 4  obj[1].y = 6

Note: normal initialization is to 0, but different values were used to illustrate the point.

Answer 4

I don't think you can assign a structure value to a structure variable. Think of obj[0] is some address of memory.

But you can do it during array of structure initialization:

test obj[2] = {{0, 0}, {1, 1}};