strcpy的功能如下:
原型声明:char *strcpy(char* dest, const char *src);
头文件:#include <string.h> 和 #include <stdio.h>
说明:src和dest所指内存区域不可以重叠且dest必须有足够的空间来容纳src的字符串。
返回指向dest的指针。
例子1:
#include<stdio.h> #include<string.h> int main(void) { char* str1="hello1"; char* str2="hello2"; printf("str1=%s ",str1); printf("str2=%s ",str2); strcpy(str1,str2); printf("str1=%s ",str1); printf("str2=%s ",str2); return 0; }
运行结果:
len@DESKTOP-BDP8J2M /cygdrive/e/c_study
$ gcc strcpy.c -o strcpy
len@DESKTOP-BDP8J2M /cygdrive/e/c_study
$ ./strcpy.exe
str1=hello1
str2=hello2
Segmentation fault (核心已转储)
例子2:
#include<stdio.h> #include<string.h> int main(void) { char str1[]="hello1"; char* str2="hello2"; printf("str1=%s ",str1); printf("str2=%s ",str2); strcpy(str1,str2); printf("str1=%s ",str1); printf("str2=%s ",str2); return 0; }
运行结果:
len@DESKTOP-BDP8J2M /cygdrive/e/c_study
$ gcc strcpy.c -o strcpy
len@DESKTOP-BDP8J2M /cygdrive/e/c_study
$ ./strcpy.exe
str1=hello1
str2=hello2
str1=hello2
str2=hello2
例子1和例子2的结论和随笔"静态内存"遇到的问题一样,strcpy的dest实参不可以是char*,必须是char[],即要是预先分配的字符内存空间。