Store int32 in int8 pointer

Question

In the below snippet, I define foo which is large enough to hold a 32-bit integer. However, when I try to store a 32-bit integer in foo , it does not fit.

int8_t * foo = malloc(sizeof(int32_t));
int32_t value = 300;
foo[0] = value;
printf("%i ", foo[0]); // 44

Is there some way to use up more elements to store the value instead of truncating it?

Answer 1

foo[0] has type int8_t , so it is only big enough for 1 byte.

Although adjacent bytes ( foo[1], foo[2], foo[3] ) are available, the compiler doesn't know you intend to copy all 4 bytes when working with just foo[0] .

Instead:

*((int32_t*)foo) = 300;

Which says, "Pretend that foo is the start of a 4-byte integer, then assign it the value 300" .