为什么要使用&用*传递函数的值? [英] Why use & to pass value for function with *?
问题描述
I wanted to ask the author, caf , from Understanding functions and pointers in C
我意识到我无法在同一页面上提问.
I realized I can't ask the question on that same page.
我仍然对使用&将指针的值传递给函数.我一直以为& q只是传递q的地址，它如何将q的值转换为int sqp(int * x)?
I'm still confused with using & to pass the value of pointer into a function. I always thought &q is just passing the address of q, how is it going to translate the value of q into int sqp(int * x)?
推荐答案
没有翻译".
您的功能
int sqp(int *x);
接受一个参数x，其类型为int *，即指向整数的指针".
takes one argument, x, whose type is int *, i.e. "pointer to integer".
因此，要调用它，您需要向其传递一个指向整数的指针，即整数的地址.
So, to call it you need to pass it a pointer to an integer, i.e. the address of an integer.
&前缀运算符用于计算其参数的地址，因此您可以执行以下操作:
The & prefix operator is used to compute the address of its argument, thus you can do:
int foo = 4711;
sqp(&foo);
使用变量foo的地址调用sqp()函数.
to call the sqp() function with the address of the variable foo.
这篇关于为什么要使用&用*传递函数的值?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！
