++上在C中取消引用指针？ [英] ++ on a dereferenced pointer in C?

问题描述

试图理解C指针的行为，我是通过以下（例如：code以下）有些惊讶：

Trying to understand the behaviour of pointers in C, I was a little surprised by the following (example code below):

#include <stdio.h>

void add_one_v1(int *our_var_ptr)
{
    *our_var_ptr = *our_var_ptr +1;
}

void add_one_v2(int *our_var_ptr)
{
    *our_var_ptr++;
}

int main()
{
    int testvar;

    testvar = 63;
    add_one_v1(&(testvar));         /* Try first version of the function */
    printf("%d\n", testvar);        /* Prints out 64                     */
    printf("@ %p\n\n", &(testvar));

    testvar = 63;
    add_one_v2(&(testvar));         /* Try first version of the function */
    printf("%d\n", testvar);        /* Prints 63 ?                       */
    printf("@ %p\n", &(testvar));   /* Address remains identical         */
}

/*

OUTPUT:

64
@ 0xbf84c6b0

63
@ 0xbf84c6b0

*/

究竟做了 * our_var_ptr ++ 在第二个函数语句（ add_one_v2 ）做，因为它显然不一样为 * our_var_ptr = * our_var_ptr +1 ？

What exactly does the *our_var_ptr++ statement in the second function (add_one_v2) do since it's clearly not the same as *our_var_ptr = *our_var_ptr +1?

推荐答案

由于操作precedence规则和事实，即 ++ 是一个后缀操作符， add_one_v2（）不取消引用指针，但 ++ 实际上是被应用的指针本身。但是，请记住C时，同时使用通过按值： add_one_v2（）是递增它的本地副本的指针，这将有没有影响任何存储在该地址的值。

Due to operator precedence rules and the fact that ++ is a postfix operator, add_one_v2() does dereference the pointer, but the ++ is actually being applied to the pointer itself. However, remember that C always uses pass-by-value: add_one_v2() is incrementing its local copy of the pointer, which will have no effect whatsoever on the value stored at that address.

作为测试，替换 add_one_v2（）与code的这些位，看看输出是如何受到影响：

As a test, replace add_one_v2() with these bits of code and see how the output is affected:

void add_one_v2(int *our_var_ptr)
{
    (*our_var_ptr)++;  // Now stores 64
}

void add_one_v2(int *our_var_ptr)
{
    *(our_var_ptr++);  // Increments the pointer, but this is a local
                       // copy of the pointer, so it doesn't do anything.
}

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