Question

When I compile this using GCC on linux, as I am waiting for input, the “hi” shows up. I do not want to use scanf, and want to know why the hi is showing while I am asking the user to input the name. Also when I want to printout the name of the file that was just passed, I get garbage characters. In netbeans, i get what I want. but on linux, it decides to act weirdly. please help

Code:

int main(int argc, char** argv)
{
char val[70];
        if(write(1, "Please input your name", 36)!=36)
        {
                return -1;
        }
        if(read(0, val, 36) < 0)
        {}

        if(write(1, val, 36)!=36)
        {}

printf("Yo");//THIS IS PRINTING OUT WAY BEFORE IT IS CALLED, ANY VARIABLE WITH A STRING     GETS PRINTED OUT, EVEN WITHOUT PRINTF BEING INVOKED
}

output:

Please input the file nameYo: hi
hi
???Om?0?a?Sm?  <<WHAT IS THIS? I DONT GET THIS ON NETBEANS

Answer 1

The third argument to write is the byte length of the string you’re trying to print. You have 36, but the string you provide is only 22 bytes long. Changing the code to look like the following will behave as you expected it to:

int main(int argc, char** argv)
{
    char val[70];
    if(write(1, "Please input your name", 22)!=22)
    {
        return -1;
    }
    if(read(0, val, 36) < 0)
    {}

    if(write(1, val, 36)!=36)
    {}

    printf("Yo");//THIS IS PRINTING OUT WAY BEFORE IT IS CALLED, ANY VARIABLE WITH A STRING     GETS PRINTED OUT, EVEN WITHOUT PRINTF BEING INVOKED
}

Note that you should probably look into using printf and scanf so that you wont have to worry about byte lengths so much.

That might look like this:

#include <stdio.h>

int main(int argc, char** argv)
{
    char val[70];
    printf("Please input your name");

    if(scanf("%69s", &val) == 1)
        printf(val);

    printf("Yo");
}