C printf shows result at the end

Hi.

I know that the problem is about C programming.

I have created an application that requires as input some parameters from
command line. After this my application do some computations and print
those parameters as fallow:

int main(int argc, char* argv[])
{
      fprintf(stdout,"%s %s",argv[1],argv[2]);   //print the value from
input
    
     //some computations

     return 0;
}

Why are the parameters shown in the final ? I need those parameters in
computations block.

Thanks

I didn't understand what do you wanna...
but i hope it helps you.

int main(int argc, char** argv) {
    if (argc >= 2) {
        int a = atoi(argv[1]);
        int b = atoi(argv[2]);
        fprintf(stdout,"Param1 - %s  Param2 - %s\n",argv[1],argv[2]);
        /* computation block */
        int computationResult = a + b;
        /* end of computation block */
        fprintf(stdout,"Computation result - %d\n",computationResult);
    } else {
        printf("Not enough parameters\n");
    }
    return 0;
}

Your question was not clear to me, but i think you are facing a buffered
i/o problem.

add a \n, so the system will send out the line.

fprintf(stdout,"%s %s\n",argv[1],argv[2]); 

or if you dont want a new line

fprintf(stdout,"%s %s",argv[1],argv[2]);
fflush(stdout);

cheers,
hex