getopt()函数
/* te.c */
#include <stdio.h>
#include <unistd.h>
void main(int argc, char **argv){
int c;
char optstr[]="i:t:";
while((c=getopt(argc,argv,optstr))!=-1){
switch(c){
case 'i':
printf("i flag\n");
printf("%s\n",optarg);
break;
case 't':
printf("t flag\n");
printf("%s\n",optarg);
break;
}
}
}
gcc编译运行
$gcc te.c -o a
$./a -i test -t
然后输出为
i flag
nci
./a: option requires an argument -- 't'
也就是说根本就没有进入case 't':中去
求解释？
[解决办法]
char optstr[]="i:t:";
多个：
改为char optstr[]="i:t";
意思不同
按照你的意思t后不跟参数能进入case 't':中去
[解决办法]
试试加一个分支
case '?':
printf("flag %c\n", optopt);
break;
[解决办法]
RETURN VALUE
The getopt() function shall return the next option character specified on the command line.
A colon ( ':' ) shall be returned if getopt() detects a missing argument and the first character of optstring was a colon ( ':' ).
A question mark ( '?' ) shall be returned if getopt() encounters an option character not in optstring or detects a missing argument and the first character of
optstring was not a colon ( ':' ).
Otherwise, getopt() shall return -1 when all command line options are parsed.
while ((c = getopt(argc, argv, ":abf:o:")) != -1) { switch(c) { case 'a': if (bflg) errflg++; else aflg++; break; case 'b': if (aflg) errflg++; else { bflg++; bproc(); } break; case 'f': ifile = optarg; break; case 'o': ofile = optarg; break; case ':': /* -f or -o without operand */ fprintf(stderr, "Option -%c requires an operand\n", optopt); errflg++; break; case '?': fprintf(stderr, "Unrecognized option: -%c\n", optopt); errflg++; } }