std::malloc

void* malloc( std::size_t size );

Allocates size bytes of uninitialized storage.

If allocation succeeds, returns a pointer to the lowest (first) byte in the allocated memory block that is suitably aligned for any scalar type.

If size is zero, the behavior is implementation defined (null pointer may be returned, or some non-null pointer may be returned that may not be used to access storage).

Parameters

Return value

On success, returns the pointer to the beginning of newly allocated memory. The returned pointer must be deallocated with std::free() or std::realloc().

On failure, returns a null pointer.

Notes

This function does not call constructors or initialize memory in any way. There are no ready-to-use smart pointers that could guarantee that the matching deallocation function is called. The preferred method of memory allocation in C++ is using RAII-ready functions std::make_unique, std::make_shared, container constructors, etc, and, in low-level library code, new-expression.

Example

#include <iostream>   
#include <cstdlib> 
 
int main() 
{
    int* p1 = (int*)std::malloc(4*sizeof(int));  // allocates enough for an array of 4 int
    int* p2 = (int*)std::malloc(sizeof(int[4])); // same, naming the type directly
    int* p3 = (int*)std::malloc(4*sizeof *p3);   // same, without repeating the type name
 
    if(p1) {
        for(int n=0; n<4; ++n) // populate the array
            p1[n] = n*n;
        for(int n=0; n<4; ++n) // print it back out
            std::cout << "p1[" << n << "] == " << p1[n] << '\n';
    }
 
    std::free(p1);
    std::free(p2);
    std::free(p3);
}

p1[0] == 0
p1[1] == 1
p1[2] == 4
p1[3] == 9

See also