char * 和 int * 的 C 指针算法不同 [英] C pointer arithmetic different for char * and int *

问题描述

有人可以解释为什么我可以使用以下代码移动 char * 而不是 int * 的内存位置:

Could someone explain why I can move through the memory locations for the char * but not for the int * using the following code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(void){
int *  a = malloc(48);  
char * s = malloc(27);

for (int i = 0;i<26;i++) s[i] = i+65;

int intSize = sizeof(int);
printf("%d\n",intSize); 

for(int i = 0;i<12;i++){
    a[i] = i;
    printf("%d %d %c\n",*(a+i*intSize),a[i],*(s+i));
}


return 0; 
}

推荐答案

当你声明一个变量时，例如

when you declare a variable e.g.

int* a;
char* b;

所有的偏移量最终都是按字节偏移量计算的，偏移量的大小取决于类型.

all offsets are in the end calculated as byte offsets, the size of the offset depends on the type.

so a + 1 实际上是 a + sizeof(int) 而 b + 1 实际上是 b + sizeof(字符)

so a + 1 is in reality a + sizeof(int) and b + 1 is in reality b + sizeof(char)

编译器会处理这个问题，它使代码更容易阅读，否则在遍历数组等时，你总是需要计算一个类型有多少字节

the compiler handles this, it makes it easier to read the code otherwise you would always have to calculate how many bytes a type has when looping through an array and such

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