Why does the printf fail to print Jack and George's names?

Question

I would like to know why Jack and George's names fail to print. I tried adding another member in the struct the names get printed normally, why is that ? I would really appreciate it if someone can help. Here's the code:

    #include <stdio.h>
    typedef unsigned short int u16;
    typedef unsigned char u8;
    typedef struct 
    {
     u8 name[10];
     u16 salary;
     u16 bonus;
     u16 deduction;
     //u8 x;//why does the printed name get ruined 
     //without this?
     }employee;
     void main (void)
     {
       employee arr[3]={{.name = "John"},{.name = 
        "Jack"},{.name = "George"}};
       u16 i = 0;
       u16 sum = 0; 
       for (i = 0; i < 3; i++)
{
    printf("\nPlease enter %s's Salary:",arr[i].name);
    scanf(" %d",&arr[i].salary);
    printf("\nPlease enter %s's Bonus:",arr[i].name);
    scanf(" %d",&arr[i].bonus);
    printf("\nPlease enter %s's Deduction:",arr[i].name);
    scanf(" %d",&arr[i].deduction);
    sum = sum + arr[i].salary + arr[i].bonus - arr[i].deduction;
}
    printf("\nTotal value needed is: %d",sum);
}

but you're showing the code that _works_ right? can you show what changes you made exactly / or the code which _doesn't_ work? — Jean-François Fabre, Dec 04 '18 at 21:21
You are using the wrong format specifier in `scanf` which leads to memory corruption. An easy fix is to change to reading an `int` instead of "u16" (which you shouldn't be definiing yourself anyway since the language already provides fixed width types) — M.M, Dec 04 '18 at 21:21
@M.M which format specifier should I use instead of %d then? — Nemo, Dec 04 '18 at 21:23
@Jean-FrançoisFabre the code doesn't work with " u8 x" commented — Nemo, Dec 04 '18 at 21:24
`%hu` would be correct in this case. But I would recommend reading into an `int` and then assigning that back to your struct field. — M.M, Dec 04 '18 at 21:25
[What is the format specifier for unsigned short int?](//stackoverflow.com/q/8699812) — 001, Dec 04 '18 at 21:25
@Nemo, `%d` will read into `int`. `%hu` is closest you get to `u16`, but its not guaranteed. — mvp, Dec 04 '18 at 21:26
If it is a memory corruption, then changing the 'shape' of the memory (like by adding more fields or variables on the stack, for example) sometimes causes the corruption to happen on a part of the memory that is not used, or that is used for something else (causing either no symptoms, or a different symptom) — Basya, Dec 04 '18 at 21:26
@M.M I thought that the memory widths sometimes differ from a compiler to another e.g sometimes int is 2 bytes and other times it's 4 bytes isn't this true ? — Nemo, Dec 04 '18 at 21:27
@Nemo `%hu` is defined as the specifier for reading `unsigned short` regardless of sizes, which is what the code has (it defines u16 as typedef for unsigned short) — M.M, Dec 04 '18 at 21:28
@Jean-FrançoisFabre how do I turn the warnings on ? I am using a gcc compiler. — Nemo, Dec 04 '18 at 21:38

score 3 · Answer 1 · answered Dec 04 '18 at 21:31

It's because your scanf()'s are wrong, and are overwriting members of your struct.

A decent compiler will issue a lot of warnings that are very severe:

$ gcc -Wall t.c
t.c:13:11: warning: return type of ‘main’ is not ‘int’ [-Wmain]
      void main (void)
           ^~~~
t.c: In function ‘main’:
t.c:22:14: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘u16 *’ {aka ‘short unsigned int *’} [-Wformat=]
     scanf(" %d",&arr[i].salary);
             ~^  ~~~~~~~~~~~~~~
             %hd
t.c:24:14: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘u16 *’ {aka ‘short unsigned int *’} [-Wformat=]
     scanf(" %d",&arr[i].bonus);
             ~^  ~~~~~~~~~~~~~
             %hd
t.c:26:14: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘u16 *’ {aka ‘short unsigned int *’} [-Wformat=]
     scanf(" %d",&arr[i].deduction);
             ~^  ~~~~~~~~~~~~~~~~~
             %hd

When you are calling scanf(" %d",&arr[i].salary); , the %d says to store the number you read as an int type.

However you have declared salary as a u16 type, which on your system is probably a lot smaller than an int. scanf jsut assumes you were telling the truth when you provided %d ,and stores an int in whatever the &arr[i].salary pointer points to, overwriting memory after that variable which might overwrite and trash the name array.

So either

declare your salary, bonus and deduction as int instead of u16, so it matches the %d argument you give to scanf or
Provide a proper type specifier to scanf for your u16 variables, which is %hu.

e.g.

   scanf(" %hu",&arr[i].salary);
   scanf(" %hu",&arr[i].bonus);
   scanf(" %hu",&arr[i].deduction);

Why does the printf fail to print Jack and George's names?

1 Answers1