Memset to set array elements to 0
c, memset
Solution
`b` is a pointer, so `sizeof(b)` is the size of a pointer, most likely 4 or 8 on current systems. So you're only setting the first few bytes to 0, instead of the entire array.
If you had declared `b` as an array, e.g.
int b[5];
then `sizeof(b)` would be the size of the entire array, and your `memset` would work as you expected.
Problem
I'm a newbie playing with memset and pointers. When I compile and run : ``` main(){ int a; int *b = (int *)malloc(5*sizeof(int)); memset(b,0, 5*sizeof(int)); if (b != NULL){ for(a=0;a<4;a++){ //b[a] = a*a; printf ("Value of b %u\n", b[a]); } } free(b); b = NULL; } ``` I am able to print all elements value as 0. However when I change the memset line to ``` memset(b,0, sizeof(b)); ``` I always get one of the elements with a huge number which I assumed to be the address of that element. However on trying to print both address and value at element with: ``` printf("Value of b %u and address %u\n", b[a], b+(a*sizeof(int))); ``` I get two long numbers which aren't the same. What exactly is happening? Have I used memset in the wrong way? Please tell me if I need to attach screenshots of output/ clarify my question. Thanks!