我正在尝试使用fork()
和exec()
在c ++中实现shell。我的代码如下:
#include <iostream>
#include <string>
#include <unistd.h>
#include <sstream>
#include <vector>
using namespace std;
int main(int argc, char** argv){
while(true){
cout << "minish> ";
string input = "";
getline(cin, input);
int count = 1;
int shouldCount = 1;
if(input.length() == 0) shouldCount = 0;
if(shouldCount == 1){
int didFindLetter = 0;
for(int i = 1; i < input.length()-1;i++){
if(input[i] ==' ' && didFindLetter == 1){
count++;
didFindLetter = 0;
}
else if(input[i]!=' '){
didFindLetter = 1;
}
}
}
//need to create a wordArray here...
vector<string> wordArray;
std::istringstream iss (input);
string tempWord;
for(int i = 0; i < count; i++){
iss >> tempWord;
wordArray.push_back(tempWord);
}
char* argsArray[1024];
count = 0;
for (int i = 0; i < wordArray.size(); i++) {
//strdup returns pointer to a char array copy of its parameter
argsArray[count++] = strdup(wordArray[i].c_str());
}
argsArray[count++] = (char *)NULL;
pid_t pid = fork();
if(pid == 0){
//child
execvp(argsArray[0], argsArray);
fprintf(stderr, "exec: %s\n", strerror(errno));
exit(errno);
}
else if(pid == 1){
//int waitStatus;
//pid_t terminated_child_pid = wait(&waitStatus);
wait(NULL);
}
}
return 0;
}
当我运行此代码并尝试执行单个命令时,它似乎正常工作
Elliots-MacBook-Pro:迷你Elliot $ ./minish
minish&GT; LS
minish&GT; makefile minish minish.cpp test
通过execv运行ls之后,代码不会打印“minish&gt;”来提示我输入更多内容,但是,如果我继续输入命令,即“ls”,则能够继续执行。
这个问题的原因是什么,我怎么能修复它?
答案 0 :(得分:2)
我不太确定我是否遵循了您的示例,但您的语义为fork()
错误:
RETURN VALUE
On success, the PID of the child process is returned
in the parent, and 0 is returned in the child. On
failure, -1 is returned in the parent, no child
process is created, and errno is set appropriately.
您的代码让父母检查返回值1
- 这意味着它会跳过等待孩子;它会立即继续进行另一次迭代。
答案 1 :(得分:0)
您的问题是,在子进程设法完成自己的输出之前,您正在获得minish
提示。
您可以通过wait
子进程来阻止此操作。子进程永远不会有pid == 1
或0
,因此您的else子句应该是
else if(pid >= 0) /*instead of `if pid==1`*/ {
wait(NULL);
}else { /*report fork failure */; }