当我在 printf() 中的字符串后添加一个带加号的 int 时会发生什么 [英] What happens when I adding an int with a plus sign after the string within printf()

问题描述

我已经在一个混淆的程序中阅读了如下代码.

I have read the code like the one down below in an obfuscated program.

我想知道为什么当我这样做时编译器给了我警告而不是错误.代码真正想要做什么以及为什么编译器建议我使用数组?

I wonder why the compiler gave me an warning instead of an error when I doing like this. What the code really want to do and why the compiler suggests me to use an array?

#include <stdio.h>
int main()
{
    int f = 1;
    printf("hello"+!f);
    return 0;
}

warning: adding 'int' to a string does not append to the string [-Wstring-plus-int]
printf("hello"+!f);
       ~~~~~~~^~~
note: use array indexing to silence this warning
printf("hello"+!f);
              ^
       &      [  ]

推荐答案

考虑语句 printf("hello");

此语句将字符串文字 "hello" 发送到 printf(); 函数.

This statement sends the string literal "hello" to the printf(); function.

现在让我们单独考虑代码

Lets now separately consider the code

char* a = "hello";

这将指向存储字符串文字 "hello" 的地址.

This would point to an address where the string literal "hello" is stored.

万一呢

char* a = "hello" + 1;

它将使 a 指向存储 "ello" 的地址."hello" + 1的地址，指向字符串字面量"ello"

It will make a point to an address where "ello" is stored. Address of "hello" + 1, which points to address of the string literal "ello"

将此应用于您的代码

printf("hello"+!f);

f 的值为 1.!f 的值为 0.因此，最终它将指向字符串文字 "hello" + 0 的地址，即 "hello".然后将其传递给 printf().

f has value 1. !f will have value 0. So, eventually it will point to the address of the string literal "hello" + 0, which is "hello". That is then passed to the printf().

您没有收到错误，因为它不是错误.

You are not getting an error because it is not an error.

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