[英]Can't execlp netcat
I'm trying to exec the following command from C:我正在尝试从 C 执行以下命令:
netcat 127.0.0.1 4444
This is my code:这是我的代码:
#include <unistd.h>
int main() {
execlp("/usr/bin/netcat", "127.0.0.1", "4444", 0);
}
However, I keep getting the error: Error: no ports specified for connection .但是,我不断收到错误消息: Error: no ports specified for connection 。
To answer a few questions I feel may come up:回答几个我觉得可能会出现的问题:
which netcat
gives /usr/bin/netcat
which netcat
给/usr/bin/netcat
execvp
and got the same resultsexecvp
并得到相同的结果-e /bin/bash
argument but I've omitted it for simplicity. Adding it doesn't change the results.)-e /bin/bash
参数,但为了简单起见,我省略了它。添加它不会改变结果。)The problem is, that your first argument is "4444".问题是,您的第一个参数是“4444”。 Why?
为什么?
Look at this line:看看这一行:
execlp("/usr/bin/netcat", "127.0.0.1", "4444", 0);
The first argument to execlp()
is the binary to execute. execlp()
的第一个参数是要执行的二进制文件。 The rest of the arguments will form the argv[]
-vector passed to the process.其余的参数将形成传递给进程的
argv[]
-vector。
By convention, argv[0]
should contain the name of the executable and the first "real" argument is argv[1]
, which is "4444" in your case.按照惯例,
argv[0]
应包含可执行文件的名称,第一个“真实”参数是argv[1]
,在您的情况下为“4444”。 So what you do is the equivalent on the shell of所以你所做的是在 shell 上的等价物
netcat 4444
网猫 4444
and the correct call would be:正确的调用是:
execlp("/usr/bin/netcat", "netcat", "127.0.0.1", "4444", 0);
Btw.顺便提一句。 the use of
execlp()
makes only sense when calling a binary without a full path, since it looks for the correct path itself (using the PATH environment variable). execlp()
的使用仅在调用没有完整路径的二进制文件时才有意义,因为它本身会寻找正确的路径(使用 PATH 环境变量)。 Otherwise, execl()
should be used.否则,应使用
execl()
。 But it will work anyway.但无论如何它都会起作用。
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