为什么＆QUOT; A + 1＆QUOT;和＆QUOT;＆安培A + 1＆QUOT;给出不同的结果，当＆QUOT; A＆QUOT;是一个int数组？ [英] Why do "a+1" and "&a+1" give different results when "a" is an int array?

问题描述

int main()
{
    int a[]={1,2,3,4,5,6,7,8,9,0};

    printf("a = %u , &a = %u\n",a,&a);
    printf("a+1 = %u , &a+1 = %u\n",a+1,&a+1);
}

如何和＆放大器;一个是内部间preTED

how a and &a are internally interpreted?

推荐答案

好了，一个是数组的第一个元素的地址，＆安培;一个是数组的地址，但显然它们都具有相同的地址

Well, a is the address of the first element of the array, and &a is the address of the array, but obviously they both have the same address.

然而，当添加（或减去）一个号码从一个指针，编译器需要的数据考虑的大小从而在你的情况（假定int的长度是4个字节）一个+ 1将是4比更大因为你移动鼠标指针领先的整数，但和放大器; A + 1将40更大，因为你更提前10整数的指针一个阵列

However when you add (or subtract) a number from a pointer the compiler takes the size of the data into consideration thus in your case (assuming the size of int is 4 bytes) a+1 will be bigger than a by 4 because you move the pointer one integer ahead, but &a+1 would be bigger by 40 because you more the pointer one ARRAY OF 10 INTEGERS ahead.

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