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(argv[1],"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(argv[1],"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 # @