High quality c ++ programming (How does 7.4 pointer parameter transmit memory ?)

7.4 how does the pointer Parameter Pass the memory?

If the function parameter is a pointer, do not expect this pointer to apply for dynamic memory.In Example 7-4-1, the getmemory (STR, 200) Statement of the test function does not enable STR to obtain the expected memory. STR is still null,

Why?

Void getmemory (char * P, int num)

{

P = (char *) malloc (sizeof (char) * num );

}

Void test (void)

{

Char * STR = NULL;

Getmemory (STR, 100); // STR is still null

Strcpy (STR, "hello"); // running error

}

Example-4-1 try to apply for dynamic memory with pointer Parameters

The fault lies in the getmemory function. The compiler always needs to make a temporary copy for each parameter of the function. The copy of the pointer parameter P is _ p, and the compiler makes _ p = P. If the program in the function body modifies the content of _ p, the content of parameter P is modified accordingly. This is why pointers can be used as output parameters. In this example, _ P applied for a new memory, but changed the memory address indicated by _ p, but P was not changed at all. Therefore, the getmemory function cannot output anything. In fact, each execution of getmemory will leak a piece of memory, because the memory is not released with free.

 If you have to use pointer parameters to request memory, you should use "pointer to Pointer" instead"See example 7-4-2.

Void getmemory2 (char ** P, int num)

{

* P = (char *) malloc (sizeof (char) * num );

}

Void Test2 (void)

{

Char * STR = NULL;

Getmemory2 (& STR, 100); // note that the parameter is & STR, not Str

Strcpy (STR, "hello ");

Cout <STR <Endl;

Free (STR );

}

Example-4-2 use a pointer to the pointer to request dynamic memory

Because the concept of "pointer to Pointer" is not easy to understand, we can use function return values to transmit dynamic memory.This method is simpler. See example 7-4-3.

Char * getmemory3 (INT num)

{

Char * P = (char *) malloc (sizeof (char) * num );

Return P;

}

Void test3 (void)

{

Char * STR = NULL;

STR = getmemory3 (100 );

Strcpy (STR, "hello ");

Cout <STR <Endl;

Free (STR );

}

Although it is easy to use the function return value to pass dynamic memory, some people often use the return statement wrong.It is emphasized that the return statement should not be used to return the pointer pointing to the "stack memory", because the function exists in it automatically disappears when it ends. See example 7-4-4.

Char * getstring (void)

{

Char P [] = "Hello World ";

Return P; // the compiler will give a warning

}

Void test4 (void)

{

Char * STR = NULL;

STR = getstring (); // STR content is junk

Cout <STR <Endl;

}

Example-4-4 Return Statement returns a pointer to "stack memory"

Use the debugger to track test4 step by step. After executing the STR = getstring statement, STR is no longer a null pointer, but the STR content is not "Hello World" but garbage.

What if I rewrite Example 7-4-4 to Example 7-4-5?

Char * getstring2 (void)

{

Char * P = "Hello World ";

Return P;

}

Void test5 (void)

{

Char * STR = NULL;

STR = getstring2 ();

Cout <STR <Endl;

}

Example-4-5 return statement returns a constant string

Although the function test5 runs without errors, the design concept of the function getstring2 is incorrect. Because "Hello World" in getstring2 is a constant string located in the static storage zone, it remains unchanged during the lifetime of the program. No matter when getstring2 is called, it returns the same read-only memory block.