Linux
Linux
C
C
Unix OS
Unix OS

Using write function to display pointer

I wrote the attached code with the intention to get 42 as the output but I instead get an asterisks (*) I know that 42 is the ACII value of an asterisk. Please look at my code and see something I might have missed. The program is in C using the <unistd.h> library, I have to use the unistd.h and not the stdio.h for this code.

[code]
#include <unistd.h>

void      ft_putchar(char c)
{
      write(1, &c, 1);
}

void ft_putnbr(int nbr)
{
write(1, &nbr, 1);
}

void      ft_ft(int *nbr)
{
      *nbr = 42;
}

int      main(void)
{
      int a;
      int *ptr;
      char c;

      a = 7;
      ptr = &a;
      ft_ft(ptr);
ft_putnbr(a);
      ft_putchar('\n');
}
/code]

Trevor MaselemeAsked: 2017-10-28

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LVL 40

evilrixSenior Software Engineer (Avast)Commented: 2017-10-30

Write just writes a series of bytes to stdout, which is being interpreted as a series of chars by the shell.

"write() writes up to count bytes from the buffer pointed buf to the file referred to by the file descriptor fd."

https://linux.die.net/man/2/write

You'll need to use printf.

http://en.cppreference.com/w/cpp/io/c/fprintf

"Loads the data from the given locations, converts them to character string equivalents and writes the results to a variety of sinks."

#include <unistd.h>
#include <cstdio>

void      ft_putchar(char c)
{
      write(1, &c, 1);
}

void    ft_putnbr(int nbr)
{
    printf("%d", nbr);
}

void      ft_ft(int *nbr)
{
      *nbr = 42;
}

int      main(void)
{
      int a;
      int *ptr;
      char c;

      a = 7;
      ptr = &a;
      ft_ft(ptr);
    ft_putnbr(a);
      ft_putchar('\n');
}

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https://ideone.com/lkj2mX

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