在用C单独的函数初始化一个指针 [英] Initializing a pointer in a separate function in C

问题描述

我需要做一个简单的事情，这在我以前在Java中做很多次，但我用C很卡（纯C，不是C ++）。情况是这样的：

I need to do a simple thing, which I used to do many times in Java, but I'm stuck in C (pure C, not C++). The situation looks like this:

int *a;

void initArray( int *arr )
{
    arr = malloc( sizeof( int )  * SIZE );
}

int main()
{
    initArray( a );
    // a is NULL here! what to do?!
    return 0;
}

我有一些初始化功能，这应该给定的指针分配到一些分配数据（无所谓）。我应该怎样给一个指向函数的指针，以该指针将被修改，然后就可以进一步在code（即函数调用返回后）一起使用？

I have some "initializing" function, which SHOULD assign a given pointer to some allocated data (doesn't matter). How should I give a pointer to a function in order to this pointer will be modified, and then can be used further in the code (after that function call returns)?

推荐答案

您需要调整*的指针，这意味着你需要一个指针传递给*一个。你是这样做的：

You need to adjust the *a pointer, this means you need to pass a pointer to the *a. You do that like this:

int *a;

void initArray( int **arr )
{
    *arr = malloc( sizeof( int )  * SIZE );
}

int main()
{
    initArray( &a );
    return 0;
}

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