#include "stdio.h"
int main(int argc, char *argv[])
{
int count;
printf("The command line has %d arguments: \n",argc-1);
for(count=1;count<argc;count++)
printf("%d: %s\n",count,argv[count]);
return 0;
}
这个程序我编译了下,得不到想要的结果啊
只有:
The command line has 3 arguments:
Press any key to continue
这是怎么回事啊?
int main(int argc, char *argv[])
{
int count;
printf("The command line has %d arguments: \n",argc-1);
for(count=1;count<argc;count++)
printf("%d: %s\n",count,argv[count]);
return 0;
}
这个程序我编译了下,得不到想要的结果啊
只有:
The command line has 3 arguments:
Press any key to continue
这是怎么回事啊?
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