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'). codepad.org/0PPoS9BU — Adrian Jandl
– Adrian Jandl, Commented Jul 15, 2013 at 6:50

P.P · Accepted Answer · 2013-07-15 06:51:43Z

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, 2013 at 6:51

P.P

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2 Comments

0decimal0 Over a year ago

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 Over a year ago

@VipulVerma Technically this answer should be the accepted one .

Grijesh Chauhan · Accepted Answer · 2013-07-15 07:57:48Z

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, 2013 at 7:57

Grijesh Chauhan

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answered Jul 15, 2013 at 6:50

Alexis

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1 Comment

0decimal0 Over a year ago

@VipulVerma He didn't answer why it has type int ... read other answers carefully.

Donotalo · Accepted Answer · 2013-07-15 06:51:42Z

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, 2013 at 6:51

Donotalo

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Comments

Community · Accepted Answer · 2017-05-23 12:21:04Z

0

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.

edited May 23, 2017 at 12:21

CommunityBot

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answered Jul 15, 2013 at 7:16

Rohan

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Comments

Saxtheowl · Accepted Answer · 2013-07-15 07:33:32Z

0

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++;
 }
 }

answered Jul 15, 2013 at 7:33

Saxtheowl

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