如何将指针作为函数参数返回 [英] How to return a pointer as a function parameter

问题描述

我正在尝试从函数参数返回数据指针:

I'm trying to return the data pointer from the function parameter:

bool dosomething(char *data){
    int datasize = 100;
    data = (char *)malloc(datasize);
    // here data address = 10968998
    return 1;
}

但是当我以以下方式调用该函数时，数据地址将更改为零:

but when I call the function in the following way, the data address changes to zero:

char *data = NULL;
if(dosomething(data)){
    // here data address = 0 ! (should be 10968998)
}

我在做什么错了?

推荐答案

您正在按值传递. dosomething修改其data的本地副本-调用者将永远不会看到它.

You're passing by value. dosomething modifies its local copy of data - the caller will never see that.

使用此:

bool dosomething(char **data){
    int datasize = 100;
    *data = (char *)malloc(datasize);
    return 1;
}

char *data = NULL;
if(dosomething(&data)){
}

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