Ç - （PTR = =放大器; PTR）什么是* PTR？ [英] C - (ptr = = &ptr) What is *ptr?

问题描述

这可能是一个愚蠢的问题，但我有一个小问题，用C指针的理解。更当谈到阵列。例如：

This might be a stupid question, but I have a little problem with understanding of C Pointers. Even more when it comes to arrays. For example:

char ptr[100];
ptr[0]=10;

fprintf(stderr, "&ptr: %p \n ptr: %p \n*ptr: %d\n", &ptr, ptr, *ptr);

if ( &ptr == ptr ) {
  fprintf(stderr, "Why?\n");
}

这怎么可能？ PTR'是在ADRESS＆放大器; PTR。和PTR的内容是一样的＆放大器; PTR。那么为什么是* PTR = 10 ???

How is this even possible? 'ptr' is at the adress &ptr. And the content of ptr is the same as &ptr. Then why is *ptr = 10 ???

推荐答案

数组的第一元素的地址是相同的数组本身的地址。

The address of the first element of the array is the same as the address of the array itself.

除了当它的操作数的的sizeof 或地址的＆安培; 运营商，或者是一个字符串文字被用来初始化声明另一个阵列，将被隐式转换（衰变）型T N个元素的数组的前pression键入指针T和值将是地址阵列中的第一个元素。

Except when it is the operand of the sizeof or address-of & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be implicitly converted ("decay") to type "pointer to T" and the value will be the address of the first element in the array.

如果除权pression A 的类型是T N个元素的数组，那么前pression ＆放;一个是键入指针至T N个元素的数组，或 T（*）[N] 。

If the expression a is of type "N-element array of T", then the expression &a is type "pointer to N-element array of T", or T (*)[N].

由于声明

T a[N];

那么下面的都是真实的：

then the following are all true:

Expression         Type        Decays to
----------         ----        ---------
         a         T [N]       T *
        &a         T (*)[N]    n/a
        *a         T           n/a

这位前pressions A 和＆放大器;一个都计算出相同的值的（阵列中的第一个元素的位置），但有不同的类型（分别为指针T和指针至T的阵列）。

The expressions a and &a both evaluate to the same value (the location of the first element in the array), but have different types (pointer to T and pointer to array of T, respectively).

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