指针运算得到的输出错误 [英] Pointer arithmetic getting wrong output

问题描述

在下面的程序，在这里， PTR 被宣布为一个指向整数指针，并分配数组的基址 P [] ，已被宣布为整型指针数组。假设 PTR 包含地址 9016 （假设P的起始地址为9016）前 PTR 是递增，并在 PTR ++ ，它将包含值9020（假设int接受4个字节）。

所以 PTR-P 应该给输出4即（9020-9016 = 4）。但它给输出1。为什么？

 ＃包括LT＆;＆stdio.h中GT;
诠释的main（）
{
    静态int类型的[] = {0,1,2,3,4};
    静态为int * P [] = {A，A + 1，+ 2，+ 3，+ 4};
    INT ** PTR = P;
    PTR ++;
    的printf（％d个，PTR-P）;
    返回0;
}
 

解决方案

一个指针的结果减去另一个指针的元素他们之间而不是字节数数。

  INT ** PTR = P;
PTR ++;
 

PTR 向前移动一个元素，因此 PTR - P 的 1的值。

顺便说一句，此行为与 PTR ++ 是一致的（即 PTR = P + 1，在你的榜样。

In the following program, Here ptr Has been declared as a pointer to an integer pointer and assigned the base address of the array p[], which has been declared as an array of integer pointer. Suppose ptr contains the address 9016 (suppose the starting address of p is 9016) before ptr is incremented and after ptr++, it will contain the value 9020 (suppose int takes 4 bytes).

So ptr-p should give the output as 4 i.e (9020-9016=4). But it is giving output as 1 . why?

#include<stdio.h>
int main()
{
    static int a[]={0,1,2,3,4};
    static int *p[]={a,a+1,a+2,a+3,a+4};
    int **ptr=p;
    ptr++;
    printf("%d",ptr-p);
    return 0;
}

解决方案

The result of one pointer minus another pointer is the number of elements between them, not the number of bytes.

int **ptr=p;
ptr++;

ptr moves forward one element, so ptr - p has a value of 1.

BTW, this behavior is consistent with ptr++ (which means ptr = p + 1; in your example.

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