C语言中的多重引用和取消引用 [英] Multiple Reference and Dereference in C

问题描述

有人可以为我解释多重引用和取消引用背后的概念吗?为什么以下程序将输出显示为"h"?

Can somebody clealry explain me the concept behind multiple reference and dereference ? why does the following program gives output as 'h' ?

int main()
{
char *ptr = "hello";
printf("%c\n", *&*&*ptr);

getchar();
return 0;
} 

不是this，而是产生'd'?

and not this , instead it produces 'd' ?

int main()
{
char *ptr = "hello";
printf("%c\n", *&*&ptr);

getchar();
return 0;
}

我读到连续使用'*'和'&'互相抵消，但此说明不能提供上述代码生成的两个不同输出背后的原因?

I read that consecutive use of '*' and '&' cancels each other but this explanation does not provide the reason behind two different outputs generated in above codes?

推荐答案

第一个程序产生h，因为& s和* s相互取消":取消引用X的地址"会返回X:

The first program produces h because &s and *s "cancel" each other: "dereferencing an address of X" gives back the X:

• ptr-指向"hello"文字的初始字符的指针
• *ptr-取消对指向初始字符(即初始字符)的指针的引用
• &*ptr指向初始字符的指针(即，指向初始字符的指针，即ptr本身)的取消引用的地址

• ptr - a pointer to the initial character of "hello" literal
• *ptr - dereference of a pointer to the initial character, i.e. the initial character
• &*ptr the address of the dereference of a pointer to the initial character, i.e. a pointer to the initial character, i.e. ptr itself

以此类推.如您所见，一对*&将您带回到开始的地方，因此您可以从取消引用/获取地址表达式中消除所有这些对.因此，您第一个程序的printf等效于

And so on. As you can see, a pair *& brings you back to where you have started, so you can eliminate all such pairs from your dereference / take address expressions. Therefore, your first program's printf is equivalent to

printf("%c\n", *ptr);

第二个程序具有未定义的行为，因为正在使用%c格式说明符将指针传递给printf.如果将相同的表达式传递给%s，则会显示单词hello:

The second program has undefined behavior, because a pointer is being passed to printf with the format specifier of %c. If you pass the same expression to %s, the word hello would be printed:

printf("%s\n", *&*&ptr);

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