在c函数中分配数组 [英] allocating an array inside a c function

问题描述

我正在尝试在函数内部分配和初始化数组，但返回后似乎无法获取值.

I'm trying to allocate and initialize an array inside a function, but I can't seem to fetch the values after returning.

这是我最后一次几乎可以正常工作的尝试

This was my last almost-working attempt

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int func(int **thing);

int main() {

 int *thing;

 func(&thing);

 printf("%d %d", thing[0], thing[1]);
}

int func(int **thing) {

 *thing = calloc(2, sizeof(int));

 *thing[0] = 1;
 *thing[1] = 2; 

 printf("func - %d %d \n", *thing[0], *thing[1]);
}

，但函数外部打印的值为1和0. 有很多关于指针的文档，但是我还没有发现涉及此特定案例的内容.关于我在做什么错的任何提示吗?

but the values printed outside the function are 1 and 0. There are lots of documentation on pointers out there, but I haven't found this specific case covered. Any tips on what I'm doing wrong ?

推荐答案

与其传递指针到指针，不如从函数中返回新分配的数组，会更容易:

Rather than passing a pointer-to-pointer, you may find it easier to return the newly allocated array from your function:

int *func();

int main() {

 int *thing;

 thing = func();

 printf("%d %d", thing[0], thing[1]);
}

int *func() {

 int *thing;

 thing = calloc(2, sizeof(int));

 thing[0] = 1;
 thing[1] = 2; 

 printf("func - %d %d \n", thing[0], thing[1]);

 return thing;
}

您的代码不起作用的原因是因为:

The reason your code doesn't work is because of this:

*thing[0]

由于运算符的优先级，您应该使用:

Due to the precedence of operators, you should use:

(*thing)[0]

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