Why isn't this printf statement executed

Question

#include<stdio.h>
#include<sys/wait.h>
#include<unistd.h>
#include<stdlib.h>

int main( int argc, char *argv[] ) {
printf("Hello, to the Simulation world,\n");

int pid = fork();
if(pid < 0) {
    fprintf(stderr,"Oops something went wrong\n");
}else if (pid == 0){
    printf("I'm child, and I do all the work, \n");
    printf("This isn't printed");
    // Any printf statement over here isn't executed \\

    execvp(argv[1], argv);
} else {
    wait(&pid);
    printf("I'm Parent and I do nothing, but I will wait till my child dies.\n");
}
return 0;
}

Any printf statement below the else if printf statement isn't shown,

Compiler used: gcc version 6.3.1 20170306 (GCC)

OS: Linux based,

My assumption: stderr is replacing current line output with its own data, but whenever input is in its own line like, printf("This Line is Printed\\n") , I can see the output. But if I write it in this way printf("This Line isn't Printed") .

or

Is it happening only for me? Or some rendering problem with OS.

Output for my above program:

Answer 1

Try using fflush(NULL); before execvp(argv[1], argv);

From man fflush :

For output streams, fflush() forces a write of all user-space buffered data for the given output or update stream via the stream's underlying write function.