我试图运行从高级Linux编程书的例子(清单3.4,第51页):为什么这个Linux编程C示例代码失败?
#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>
/* Spawn a child process running a new program. PROGRAM is the name
of the program to run; the path will be searched for this program.
ARG_LIST is a NULL-terminated list of character strings to be
passed as the program’s argument list. Returns the process ID of
the spawned process. */
int spawn(char* program, char** arg_list) {
pid_t child_pid;
/* Duplicate this process. */
child_pid = fork();
if (child_pid != 0)
/* This is the parent process. */
return child_pid;
else {
/* Now execute PROGRAM, searching for it in the path. */
execvp(program, arg_list);
/* The execvp function returns only if an error occurs. */
fprintf(stderr, "an error occurred in execvp\n");
abort();
}
return 0;
}
int main() {
/* The argument list to pass to the "ls” command. */
char* arg_list[] = { "ls", /* argv[0], the name of the program. */
"-l", "/", NULL /* The argument list must end with a NULL. */
};
/* Spawn a child process running the "ls” command. Ignore the
returned child process ID. */
spawn(" ls", arg_list);
printf("done with main program\n");
return 0;
}
而且我得到了:
an error occurred in execvp
done with main program
任何想法有什么不对吗? (使用Ubuntu 10.10)
'ls'之前的空间可以成为问题吗? –
尝试''ls“'而不是''ls”'。甚至是“/ bin/ls”。此外,你可以从'errno'获得关于错误的信息;你为什么不检查那个? –
而不是(或除了)fprintf,添加'perror(“Execvp失败”);'。这将打印出一个人类可读的错误消息,解释发生了什么。 – Neal