Dynamically allocating memory for const char string using malloc()

问题

I am writing a program that reads a value from an .ini file, then passes the value into a function that accepts a PCSTR (i.e. const char *). The function is getaddrinfo().

So, I want to write PCSTR ReadFromIni(). To return a constant string, I plan on allocating memory using malloc() and casting the memory to a constant string. I will be able to get the exact number of characters that were read from the .ini file.

Is that technique okay? I don't really know what else to do.

The following example runs fine in Visual Studio 2013, and prints out "hello" as desired.

const char * m()
{
    char * c = (char *)malloc(6 * sizeof(char));
    c = "hello";
    return (const char *)c;
}    

int main(int argc, char * argv[])
{
    const char * d = m();
    std::cout << d; // use PCSTR
}

回答1:

The second line is "horribly" wrong:

char* c = (char*)malloc(6*sizeof(char));
// 'c' is set to point to a piece of allocated memory (typically located in the heap)
c = "hello";
// 'c' is set to point to a constant string (typically located in the code-section or in the data-section)

You are assigning variable c twice, so obviously, the first assignment has no meaning. It's like writing:

int i = 5;
i = 6;

On top of that, you "lose" the address of the allocated memory, so you will not be able to release it later.

You can change this function as follows:

char* m()
{
    const char* s = "hello";
    char* c = (char*)malloc(strlen(s)+1);
    strcpy(c,s);
    return c;
}

Keep in mind that whoever calls char* p = m(), will also have to call free(p) at some later point...

回答2:

One way is to return a local static pointer.

const char * m()
{
    static char * c = NULL;
    free(c);

    c = malloc(6 * sizeof(char));
    strcpy(c, "hello"); /* or fill in c by any other way */
    return c;
}

This way, whenever the next time m() is called, c still points to the memory block allocated earlier. You could reallocate the memory on demand, fill in new content, and return the address.

回答3:

NO. This is not OK. When you do

c = "hello";

the memory allocated by malloc lost.
You can do as

const char * m()
{
    char * c = (char *)malloc(6 * sizeof(char));
    fgets(c, 6, stdin);
    return (const char *)c;
}

来源：https://stackoverflow.com/questions/21579819/dynamically-allocating-memory-for-const-char-string-using-malloc

标签

c++

malloc

const-char

const-pointer