字符串指针 整形指针

字符串指针

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void tmp(char *str)
{    
    printf("2222 %p,%p\n",&str,str);
    str = "123456";
    printf("3333 %p,%p\n",&str,str);
}

void main()
{
    char *a=malloc(sizeof(char)*6);
    printf("1111 %p,%p\n",&a,a);
    tmp(a);
    printf("%s\n",a);
    printf("Done\n");
}

结果：

1111 0xbfcfe26c,0x8cce008
2222 0xbfcfe250,0x8cce008
3333 0xbfcfe250,0x80485cc

Done

整型指针：

#include <stdio.h>
#include <stdlib.h>
void tmp(int *nu,char *ch)
{
    printf("2222 &nu is %p,nu is %p,&ch is %p,ch is %p\n",&nu,nu,&ch,ch);
    *nu = 15;
    *ch = 'a';
    printf("3333 &nu is %p,nu is %p,&ch is %p,ch is %p\n",&nu,nu,&ch,ch);
}

int main()
{
    int *a = malloc(sizeof(int));
    char *c = malloc(sizeof(char));
    printf("1111 &a  is %p,a  is %p,&c  is %p,c  is %p\n",&a,a,&c,c);
    tmp(a,c);
    printf("a is%d,c is %c\n",*a,*c);
    printf("Done\n");
}

结果：

1111 &a  is 0xbfd32518,a  is 0x83c9008,&c  is 0xbfd3251c,c  is 0x83c9018
2222 &nu is 0xbfd324f0,nu is 0x83c9008,&ch is 0xbfd324f4,ch is 0x83c9018
3333 &nu is 0xbfd324f0,nu is 0x83c9008,&ch is 0xbfd324f4,ch is 0x83c9018
a is15,c is a
Done