Synopsis

Allocating memory with malloc without using sizeof.

Enabled by default

Yes

Severity/Certainty

Low/Medium

Full description

Memory was allocated with malloc() but the sizeof operator might not have been used. Using sizeof when allocating memory avoids any machine variations in the sizes of data types, and consequently avoids under-allocating. To pass this check, assign the address of the allocated memory to a char pointer, because sizeof(char) always returns 1.

Coding standards

Code examples

The following code example fails the check and will give a warning:

#include <stdlib.h>

void example(void) {
	int *x = malloc(4);  //no sizeof in malloc call
	free(x);
}

The following code example passes the check and will not give a warning about this issue:

#include <stdlib.h>

void example(void) {
	int *x = malloc(sizeof(int));
	free(x);
}