如何越过一个指针和堆分配空间 [英] How to pass over a pointer and allocate space on heap

问题描述

我想传递一个字符指针的函数，然后用东西填满它，以便调用函数可以使用它即可。因为它总是给我怪异的东西在调用函数中我写了什么，我做了简单的重新presentation：

I am trying to deliver a char pointer to a function which then fills it with stuff so that the calling function could use it then. As it always gave me weird stuff in the calling function i wrote a simple representation about what I have done:

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

void bla(char *t)
{
    t = (char*) malloc(5);
    if (t != 0) {
        t[0] = 'h';
        printf("%c\n", t[0]);
    }
}

int main(int argc, char **argv)
{
    char b[10];

    bla(b);
    printf("%c\n", b[0]);
    return 1;
}

我不太确定是否有事可做了C经过论证的副本。我需要一个指针传递给一个指针则还是有一个更好的解决方案？

I'm not quite sure if it has something to do that C passes a copy of the argument. Do I need to pass a pointer to a pointer then or is there a better solution?

编辑：

对不起球员，但我没有得到它。能否请您过目这个例子：

Sorry guys but I didn't get it. Could you please look over this example:

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

void blub(char **t)
{
    printf("%d\n", *t);
    *t = (char*) malloc(sizeof(char) * 5);
    *t[0] = 'a';
    *t[1] = 'b';
    printf("%d\n", *t);
    printf("%d\n", *t[0]);
    printf("%d\n", *t[1]);
}

int main(int argc, char **argv)
{
    char *a;
    blub(&a);
    printf("%d\n", a);
    printf("%d\n", a[0]);
    printf("%d\n", a[1]);

    return 1;
}

输出如下所示：

./main
6154128
140488712
97
98
140488712
97
0      <== THIS SHOULD BE 98 AS ABOVE!?

为什么我得到的功能blafu 98和主要的它是一个空指针？我完全以困惑：/

Why do i get 98 in the function blafu and in main it is a null pointer?! I am totaly confused :/

推荐答案

您需要使用指针的指针。

You need to use pointer to pointer.

//............vv
void bla(char **t)

然后，通过提领它使用指针：

Then use the pointer by dereferencing it:

// note the '*' in front of t
*t = (char*) malloc(5);
if (*t != 0) {
    *t[0] = 'h';
    printf("%c\n", *t[0]);
}

此外，声明 B 为的char * ，而不是字符B〔10] 。

为什么呢？因为你想改变指针。其中的逻辑是一样的其他类型，但它有点困惑在这里。结果
想想看，是这样的：如果你需要传递一个 INT ，你需要去改变它，你需要一个指针为int 。同样是在这里 - 你需要传递的字符指针并需要哟改变它，所以用指针的字符指针的 ：）

Why? Because you're trying to change the pointer. The logic is the same as other types, but it's kinda confusing here.
Think about it like this: if you need to pass an int and you need to change it, you need a pointer to int. The same is here - you need to pass a pointer to char and you need yo change it, so use "pointer to pointer to char" :)

的 修改的：
根据你的编辑 - 是的，你明白完美这一点，但只是一个小问题 - 运算符* 和优先运算符[] 。如果更换 * T [X] 与（* T）[X] ，everythng将被罚款:)

EDIT: According to your edit - yes, you have understood this perfectly, there's just a small problem - the priority of operator* and operator[]. If you replace your *t[X] with (*t)[X], everythng will be fine :)

这篇关于如何越过一个指针和堆分配空间的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！