C语言完美演绎9-23

/* 范例：9-23 */

#include <stdio.h>

#include <stdlib.h>

void main(int argc,char *argv\[\])

{

FILE *fp1;

int pos;

if((fp1=fopen(argv1,"w"))==NULL)

{

printf("File Open Error!\n");

exit(1);

}

fputs("abcdefg",fp1);

fputs("\n0123456",fp1);

fclose(fp1);

/* 可擦写(更新) */

if((fp1=fopen(argv1,"r+"))==NULL)

{

printf("文件打开错误\n");

exit(1);

}

pos = fseek(fp1,4,0); /* 从文件头跳4bytes */

putc(65,fp1);

printf("pos = %d\n",pos);

fseek(fp1,6,2); /* 从文件尾跳6bytes */

putc('#',fp1);

fseek(fp1,3,1); /* 从目前位置跳3bytes */

putc('@',fp1);

fclose(fp1);

}

程序执行结果：(假设输出文件名out)

D:\TC>p9-23 out

pos = 0

D:\TC>type out => 检查文件out内容

abcdAfg

0123456 # @