Xim*_*mik 7 c unix linux posix
我的测试应用程序是
#include <sys/types.h>
#include <sys/wait.h>
#include <signal.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <fcntl.h>
int main(int argc, char *argv[], char *envp[]) {
int fd[2];
if(pipe(fd) < 0) {
printf("Can\'t create pipe\n");
exit(-1);
}
pid_t fpid = fork();
if (fpid == 0) {
close(0);
close(fd[1]);
char *s = (char *) malloc(sizeof(char));
while(1) if (read(fd[0], s, 1)) printf("%i\n", *s);
}
close(fd[0]);
char *c = (char *) malloc(sizeof(char));
while (1) {
if (read(0, c, 1) > 0) write(fd[1], c, 1);
}
return 0;
}
Run Code Online (Sandbox Code Playgroud)
我希望在每次输入char后看到char-code.但实际上*s仅在控制台中的"\n"之后打印.所以似乎stdin(带有desc 0的文件)被缓冲了.但是read函数是无缓冲的,不是吗?哪里错了.
UPD:我使用linux.
所以解决方案是
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <termios.h>
int main(int argc, char *argv[], char *envp[]) {
int fd[2];
if(pipe(fd) < 0) {
printf("Can\'t create pipe\n");
exit(-1);
}
struct termios term, term_orig;
if(tcgetattr(0, &term_orig)) {
printf("tcgetattr failed\n");
exit(-1);
}
term = term_orig;
term.c_lflag &= ~ICANON;
term.c_lflag |= ECHO;
term.c_cc[VMIN] = 0;
term.c_cc[VTIME] = 0;
if (tcsetattr(0, TCSANOW, &term)) {
printf("tcsetattr failed\n");
exit(-1);
}
pid_t fpid = fork();
if (fpid == 0) {
close(0);
close(fd[1]);
char *s = (char *) malloc(sizeof(char));
while(1) if (read(fd[0], s, 1)) printf("%i\n", *s);
}
close(fd[0]);
char *c = (char *) malloc(sizeof(char));
while (1) {
if (read(0, c, 1) > 0) write(fd[1], c, 1);
}
return 0;
}
Run Code Online (Sandbox Code Playgroud)
在我看来,您的解决方案有点复杂。还是不明白为什么需要管道和2个进程。
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <termios.h>
int main(int argc, char *argv[], char *envp[]) {
struct termios term, term_orig;
if(tcgetattr(0, &term_orig)) {
printf("tcgetattr failed\n");
exit(-1);
}
term = term_orig;
term.c_lflag &= ~ICANON;
term.c_lflag |= ECHO;
term.c_cc[VMIN] = 0;
term.c_cc[VTIME] = 0;
if (tcsetattr(0, TCSANOW, &term)) {
printf("tcsetattr failed\n");
exit(-1);
}
char ch;
while (1) {
if (read(0, &ch, 1) > 0)
printf(" %d\n", ch);
}
return 0;
}
Run Code Online (Sandbox Code Playgroud)