在 C 中的单独函数中初始化指针 [英] Initializing a pointer in a separate function in C
问题描述
我需要做一件简单的事情,我曾经在 Java 中做过很多次,但我被困在 C(纯 C,而不是 C++)中.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"函数,它应该为一些分配的数据分配一个给定的指针(无关紧要).我应该如何给出一个指向函数的指针,以便修改这个指针,然后可以在代码中进一步使用(在函数调用返回之后)?
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
你需要调整*a 指针,这意味着你需要传递一个指针给*a.你这样做:
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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