【发布时间】：2015-12-18 15:26:56
【问题描述】：
我正在制作自己的 c shell。 shell 有一个历史功能，可以显示过去的命令。我已经做到了这一点，但是我无法重新运行旧命令。当用户输入 rr 时，我试图重新运行最后一个命令。 当 rr 为 = args[0] 时，在 main 函数中标记了问题。
这是我的代码：
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#define MAX_LINE 80 /* 80 chars per line, per command, should be enough. */
#define HIST_SIZE 12
int place=0;
char cmd[MAX_LINE+1]; //used to keep copy of inBuffer to put in *hist
/** The setup() routine reads in the next command line string storing it in the input buffer.
The line is separated into distinct tokens using whitespace as delimiters. Setup also
modifies the args parameter so that it holds points to the null-terminated strings which
are the tokens in the most recent user command line as well as a NULL pointer, indicating the
end of the argument list, which comes after the string pointers that have been assigned to
args. ***/
void setup(char inputBuffer[], char *args[],int *background){
int length, /* # characters in the command line */
start, /* Beginning of next command parameter */
i, /* Index for inputBuffer arrray */
j; /* Where to place the next parameter into args[] */
/* Read what the user enters */
length = read(STDIN_FILENO, inputBuffer, MAX_LINE);
strcpy(cmd,inputBuffer);
start = -1;
j = 0;
if (length == 0)
exit(0); /* Cntrl-d was entered, end of user command stream */
if (length < 0){
perror("error reading command");
exit(-1); /* Terminate with error code of -1 */
}
/* Examine every character in the input buffer */
for (i = 0; i < length; i++) {
switch (inputBuffer[i]){
case ' ':
case '\t' : /* Argument separators */
if(start != -1){
args[j] = &inputBuffer[start]; /* Set up pointer */
j++;
}
inputBuffer[i] = '\0'; /* Add a null char; make a C string */
start = -1;
break;
case '\n': /* Final char examined */
if (start != -1){
args[j] = &inputBuffer[start];
j++;
}
inputBuffer[i] = '\0';
args[j] = NULL; /* No more arguments to this command */
break;
case '&':
*background = 1;
inputBuffer[i] = '\0';
break;
default : /* Some other character */
if (start == -1)
start = i;
}
}
args[j] = NULL; /* Just in case the input line was > 80 */
}
int displayHistory(char *hist[], int place){
int i=place;
int hist_num=1;
do{
if(hist[i]){
printf("%d %s", hist_num, hist[i]);
hist_num++;
}
i=(i+1)%HIST_SIZE;
} while(i != place);
return 0;
}
int main(void){
char inputBuffer[MAX_LINE]; /* Input buffer to hold the command entered */
char *args[MAX_LINE/2+1];/* Command line arguments */
int background; /* Equals 1 if a command is followed by '&', else 0 */
char *hist[HIST_SIZE];
int i=0;
for(i=0; i<HIST_SIZE; i++){
hist[i]=NULL;
}
while (1){ /* Program terminates normally inside setup */
background = 0;
printf("COMMAND--> "); /* Shell prompt */
fflush(0);
setup(inputBuffer, args, &background); /* Get next command */
if(strcmp(args[0],"rr")==0){
//rr does not go into *hist
}
else{
cmd[strlen(cmd)]='\0';
free(hist[place]);
hist[place]=strdup(cmd);
place=(place+1)%HIST_SIZE;
}
/* Fill in the code for these steps:
(1) Fork a child process using fork(),
(2) The child process will invoke execvp(),
(3) If background == 0, the parent will wait,
o/w returns to the setup() function. */
pid_t pid=fork();
int status;
if(pid<0){ //error forking
fprintf(stderr, "Fork Failed");
exit(-1);
}
else if(pid==0){ //child process
if(strcmp(inputBuffer,"h")==0){
displayHistory(hist,place);
}
else if(strcmp(args[0],"rr")==0){
printf("%s",hist[place-1]); //this is here to show that hist[place-1] does have the last command
execvp(hist[place-1], args); //PROBLEM HERE. why doesn't the hist[place-1]get executed?
}
else{
if(execvp(inputBuffer, args)<0){
printf("Invalid Command\n");
}
}
}
else{ //parent process
if(background==0){ //wait
waitpid(pid, &status, 0);
}
}
memset(inputBuffer,0,sizeof(inputBuffer)); //this is done so garbage chars from old input does not show up later when history is displayed
memset(cmd, 0, sizeof(cmd));
}
}
解释为什么不执行旧命令会很好。真正让命令运行的方法会很棒。
【问题讨论】：
-
cmd[strlen(cmd)]='\0';应该达到什么目的？ -
@WeatherVane 有助于使字符串具有空终止符
-
如果没有字符串终止符，
strlen将首先不起作用。该指令将'\0'写入已经存在'\0'的确切位置 -
应该是
hist[(current+HIST_SIZE-1) % HIST_SIZE]？ -
评论太多，所以我将其作为答案。