如何对malloc函数内，用C返回指针？ [英] How to malloc inside a function and return pointer in C?

问题描述

下面是一些psudo，但我试图做到这一点。问题是书面，它返回一个空指针。

Below is some psudo, but I'm trying to accomplish this. The problem is as written, it returns a blank pointer.

int testFunction(char *t) {
    int size = 100;
    t = malloc(100 + 1);
    t = <do a bunch of stuff to assign a value>;
    return size;
}

int runIt() {
    char *str = 0;
    int str_size = 0;
    str_size = testFunction(str);
    <at this point, str is blank and unmodified, what's wrong?>
    free(str);
    return 0;
}

这工作正常，如果我有一个predefined大小，如字符海峡[100] =的，我不尝试的malloc或免费的内存后记。我需要能够使大小动态虽然。

This works fine if I have a predefined size, such as char str[100] = "" and I don't try to malloc or free memory afterwords. I need to be able to make the size dynamic though.

我也试了这一点，但似乎不知何故碰到一个腐败的指针。

I've also tried this, but seem to run into a corrupt pointer somehow.

int testFunction(char **t) {
    int size = 100;
    t = malloc(100 + 1);
    t = <do a bunch of stuff to assign a value>;
    return size;
}

int runIt() {
    char *str = 0;
    int str_size = 0;
    str_size = testFunction(&str);
    <at this point, str is blank and unmodified, what's wrong?>
    free(str);
    return 0;
}

谢谢！

推荐答案

您快到了与第二个例子，但变化

You're nearly there with the second example, but change

int testFunction(char **t) {
  ...
  t = malloc(100 + 1);

要

int testFunction(char **t) {
  ...
  *t = malloc(100 + 1);

问题的关键是，你要传递一个的char ** ，一个指针的指针，所以你要到的malloc分配到什么点是：（a指针）。

The point being that you're passing in a char**, a pointer to a pointer, so you want to assign the malloc to what that points at (a pointer).

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