使用 printf 时，C fork 是 parent 的副本

Question

#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>

int main(void) {
    write(STDOUT_FILENO,"1",1);
    if(fork() > 0) {
        write(STDOUT_FILENO,"2",1);
        write(STDOUT_FILENO,"3",1);
    }
    else {
        write(STDOUT_FILENO,"4",1);
        write(STDOUT_FILENO,"5",1);
    }
    write(STDOUT_FILENO,"\n",1);
    return 0;
}

输出为1 2 3 \n 4 5 \n

为什么如果我替换 printf（最后没有换行符）的所有写入函数，如 write(STDOUT_FILENO,"1",1)==printf("1")，我得到 1 2 3 \n 1 4 5 \n，如 child正在复制叉子上方的行？

Answer 1

是的，这是因为 stdout 流被缓冲，直到它按照这个 post Why does printf not flush after the call unless a newline is in the format string?. So when you form a new process this buffer is copied to the child's stdout buffer since we are essentially making a new memory space for the child. (Actually not quite, see Specifically, how does fork() handle dynamically allocated memory from malloc() in Linux? )

这将起作用

#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
int main(void) {

printf("1");
fflush(stdout);
if(fork() > 0) {
        printf("2");
        printf("3");
} else {
        printf("4");
        printf("5");

}
printf("\n");
return 0;
}

使用 printf 时，C fork 是 parent 的副本

C fork is a copy of parent when using printf

c

printf

fork

parent