2-D character array

Question

#include<stdio.h>
int main()
{
char str[3][10]={
                    "vipul",
                    "ss",
                    "shreya"
};

Why this won't work:

printf("%s",str[1][0]);

If i want to access str whereas

printf("%s",&str[1][0]);

or this would do it perfectly

printf("%s",str[1]);

Can anyone explain ? Why is the first code giving an error

prog.c: In function ‘main’:
prog.c:9:5: error: format ‘%s’ expects argument of type ‘char *’, but 
                   argument 2 has type ‘int’ [-   Werror=format]
cc1: all warnings being treated as errors

Why does the argument has type int?

That would be the same as saying `printf("%s", 'a')`. http://codepad.org/0PPoS9BU — Adrian Jandl, Jul 15 '13 at 06:50

score 3 · Answer 1 · edited Jul 15 '13 at 07:57

3

Well

str[1] is a char* and str[1][0] is a char.
But when you use %s, printf() expect a pointer so you try to cast the char into a pointer.

So your char is promoted to an int.

edited Jul 15 '13 at 07:57

Grijesh Chauhan

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answered Jul 15 '13 at 06:50

Alexis

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@VipulVerma He didn't answer why it has type `int` ... read other answers carefully. – 0decimal0 Jul 15 '13 at 07:00

score 3 · Accepted Answer · answered Jul 15 '13 at 06:51

3

printf("%s",str[1][0]);

The problem is in this line. When For %s format specifier, printf() expects a pointer to a null terminated string. Whereas str[1][0] is simply a char (specifically the first s in "ss"), which is promoted to int (default argument promotions). That's exactly what the error message says.

answered Jul 15 '13 at 06:51

P.P

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Strange that the answer which doesn't answer that why argument has type `int` got the most votes. +1 for you and all similar answers. – 0decimal0 Jul 15 '13 at 06:59
@VipulVerma Technically this answer should be the accepted one . – 0decimal0 Jul 15 '13 at 07:02

score 1 · Answer 3 · answered Jul 15 '13 at 06:51

1

It is said in the error:

format ‘%s’ expects argument of type ‘char *’

and your argument str[1][0] is a char, not the expected char *. In C, a char is treated as an int.

answered Jul 15 '13 at 06:51

Donotalo

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score 0 · Answer 4 · edited May 23 '17 at 12:21

In first case printf("%s",str1[1][0]); you are passing single character to printf function and format specifier that you use it %s. For %s printf function expects string of character and not character.So it gives error.
As in 1st printf function you are specifying %s and you are passing character, argument promotion will takes place and char will promoted to int.

•The default argument promotions are char and short to int/unsigned int and float to double
•The optional arguments to variadic functions (like printf) are subject to the default argument promotions

More on Default argument promotion and here.

score 0 · Answer 5 · answered Jul 15 '13 at 07:33

on your line error:

 printf("%s",str[1][0]);

you try to print a string where you have a character ("%c" in printf)

so to only print one of ur 2D array you would have to do something like that:

  int main()
  {
  int i;
  char str[3][10]=
  {
  "vipul",
  "ss",
  "shreya"
  };
  i = 0;
  while(str[0][i] != '\0')
  {
  printf("%c",str[0][i]);
  i++;
  }
  }

which is pretty ugly ^^

instead you could print all ur 2D array with 3 single iteration with that:

 int main()
 {
 int i;
 char str[3][10]=
 {
 "vipul",
 "ss",
 "shreya"
 };
 i = 0;
 while(i < 3)
 {
 printf("%s\n",str[i]);
 i++;
 }
 }

2-D character array

5 Answers5