如何更改作为参数传递的变量的值? [英] How to change value of variable passed as argument?

问题描述

如何更改在 C 中作为参数传递的变量的值?我试过这个:

How to change value of variable passed as argument in C? I tried this:

void foo(char *foo, int baa){
    if(baa) {
        foo = "ab";
    } else {
        foo = "cb";
    }
}

并调用:

char *x = "baa";
foo(x, 1);
printf("%s
", x);

但它打印 baa 为什么?提前致谢.

but it prints baa why? thanks in advance.

推荐答案

您想要更改 char* 指向的位置，因此您需要在 foo() 带有一级间接；一个 char**(指向 char 指针的指针).

You're wanting to change where a char* points, therefore you're going to need to accept an argument in foo() with one more level of indirection; a char** (pointer to a char pointer).

因此 foo() 将被改写为:

void foo(char **foo /* changed */, int baa)
{
   if(baa) 
   {
      *foo = "ab"; /* changed */
   }
   else 
   {
      *foo = "cb"; /* changed */
   }
}

现在，当调用 foo() 时，您将使用地址运算符 (&):

Now when calling foo(), you'll pass a pointer to x using the address-of operator (&):

foo(&x, 1);

您的错误代码段打印 baa 的原因是因为您只是为 local 变量 char *foo 分配了一个新值，与 x 无关.因此 x 的值永远不会被修改.

The reason why your incorrect snippet prints baa is because you're simply assigning a new value to the local variable char *foo, which is unrelated to x. Therefore the value of x is never modified.

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