Linux 将参数传递给系统调用
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Passing parameters to system calls
提问by Jeff
I did a basic helloWorld system call example that had no parameters and was just:
我做了一个基本的 helloWorld 系统调用示例，它没有参数，只是：
int main()
{
syscall(__NR_helloWorld);
return 0;
}
But now I am trying to figure out how to pass actual arguments to the system call (ie. a long). What is the format exactly, I tried:
但是现在我想弄清楚如何将实际参数传递给系统调用（即 a long）。究竟是什么格式，我试过：
int main()
{
long input = 1;
long result = syscall(__NR_someSysCall, long input, long);
return 0;
}
Where it takes a longand returns a long, but it is not compiling correctly; what is the correct syntax?
它需要 along并返回 a long，但它没有正确编译；什么是正确的语法？
采纳答案by Nikolai Fetissov
Remove the type names. It's just a function call. Example:
删除类型名称。这只是一个函数调用。例子：
#include <sys/syscall.h>
#include <unistd.h>
int main( int argc, char* argv[] ) {
char str[] = "boo\n";
syscall( __NR_write, STDOUT_FILENO, str, sizeof(str) - 1 );
return 0;
}
回答by wallyk
The prototype for syscallis
的原型syscall是
#define _GNU_SOURCE /* or _BSD_SOURCE or _SVID_SOURCE */
#include <unistd.h>
#include <sys/syscall.h> /* For SYS_xxx definitions */
int syscall(int number, ...);
This says that it takes a variable number (and type) of parameters. That depends on the particular system call. syscallis not the normal interface to a particular system call. Those can be explicitly coded, for example write(fd, buf, len).
这表示它需要可变数量（和类型）的参数。这取决于特定的系统调用。 syscall不是特定系统调用的正常接口。那些可以被显式编码，例如write(fd, buf, len)。
