我正在编写一个虚拟 shell,当用户键入 ctrl-C 时它不应该终止,而应该只生成一个新的提示行。目前,当我键入 ctrl-C 时,我的 shell 不会终止,但它仍然不会打印新的提示行。你知道为什么会这样,我该如何解决这个问题?
我的代码如下:
#include <stdlib.h>
#include <stdio.h>
#include <errno.h>
#include <string.h>
#include <unistd.h>
#include <sys/wait.h>
#include <signal.h>
#define BUFFER_SIZE 1<<16
#define ARRAY_SIZE 1<<16
void INThandler(int);
static void parseCmdArgs(char *buffer, char** cmdArgs,
size_t cmdArgsSize, size_t *nargs)
{
char *bufCmdArgs[cmdArgsSize];
char **temp;
char *buf;
size_t n, p;
cmdArgs[0] = buf = bufCmdArgs[0] = buffer;
for(temp=bufCmdArgs; (*temp=strsep(&buf, " \n\t")) != NULL ;){
if ((*temp != '\0') && (++temp >= &bufCmdArgs[cmdArgsSize]))
break;
}
for (p=n=0; bufCmdArgs[n]!=NULL; n++){
if(strlen(bufCmdArgs[n])>0)
cmdArgs[p++]=bufCmdArgs[n];
}
*nargs=p;
cmdArgs[p]=NULL;
}
void INThandler(int sig)
{
printf("\n> ");
signal(sig, SIG_IGN);
}
int main(void)
{
char buffer[BUFFER_SIZE];
char *args[ARRAY_SIZE];
int retStatus;
size_t nargs;
pid_t pid;
printf("$dummyshell\n");
signal(SIGINT, INThandler);
while(1){
printf("> ");
fgets(buffer, BUFFER_SIZE, stdin);
parseCmdArgs(buffer, args, ARRAY_SIZE, &nargs);
if (nargs==0)
continue;
if (!strcmp(args[0], "help"))
{
printf("cat cd (absolute path references only\n");
printf("exit\n");
printf("help history\n");
printf("jobs kill\n");
printf("ls more\n");
printf("ps pwd\n");
continue;
}
if (!strcmp(args[0], "exit" ))
exit(0);
pid = fork();
if (pid){
wait(&retStatus);
}
else {
if( execvp(args[0], args)) {
fprintf(stderr, "%s\n", strerror(errno));
exit(127);
}
}
/* pid = fork();
if (pid == 0)
setpgrp();
else if (pid)
pid = wait(&retStatus);
else {
if (execvp(args[0], args)){
fprintf(stderr, "%s\n", strerror(errno));
exit(127);
}
}*/
}
return 0;
}