In this section:

Overview

This error is generated whenever a function returns a pointer to a block of memory which is then ignored by the calling routine. In this case, the allocated memory block is permanently lost and can never be freed.

Code	Description	Enabled	Reported	Platform
LEAK_RETURN	Memory leaked leaving scope		Runtime	Windows/Unix

Problem

The following code calls the function gimme, which allocates a memory block that is never freed.

/*
 * File: leakscop.c
 */
#include <stdlib.h>

void gimme()
{
	char *p;
	p = malloc(10);
	return;
}

main()
{
	gimme();
	return (0);
}

Diagnosis at Runtime

[leakscop.c:10] **LEAK_SCOPE**
>>			return;
Memory leaked leaving scope: <return>
Lost block:			0x0003870 thru 0x00013879 (10 bytes)
					block allocated at:
					malloc() (interface)
					gimme() leakscop.c, 9
					main() leakscop.c, 15
Stack trace where the error occurred:
			gimme() leakscop.c,

• Line 2: Source file and line at which the problem was detected.
• Line 3: Description of the problem and the block that is to be lost.
• Line 4: Description of the block of memory that is about to be lost, including its size and the line number at which it was allocated.
• Line 10: Stack trace showing the function call sequence leading to the error.

Repair

This problem usually results from an oversight on the part of the programmer and is cured by simply freeing a block before returning from a routine. In the current example, a call to free(p) before line 10 would cure the problem.

An easy way to generate this error is to return from the middle of a routine, possibly due to an error condition arising, without freeing previously allocated data. This bug is easy to introduce when modifying existing code.

Some applications may be unable to free memory blocks and may not need to worry about their permanent loss. To suppress these error messages in this case, suppress LEAK_SCOPE in the Suppressions Control Panel.