Why does printf prints only first character of a character array whose input was read through scanf

Question

This code is supposed to read input string through scanf and store the input string in an array declared here as msg[] and then print those input string, but when I run the code all it prints is just the first character of the string. It looks really simple but I am not able to solve it.

If I type i do, it only reports i.

#include <stdio.h>

int main()
{
    char msg[5]; 
    printf("enter the message: ");
    scanf("%s",msg);           
    printf("entered msg is: %s", msg); 
}

your array can store up to `4 chars` plus a `null terminator`....... — LPs, Mar 17 '17 at 14:47
Not _at least_, but *up to* 4 chars. BTW what is the user input? — LPs, Mar 17 '17 at 14:49
I assume the input is _"more than the array can handle"_ sort? — Sourav Ghosh, Mar 17 '17 at 14:51
sorry atmost 4 characters. any input that i give it prints only first chracter for eg if i give input "i do" it prints only 'i' and yea even when i give input more than the specified size of array for eg " i love the foods that i cook" even then it prints only 'i' — hageApple, Mar 17 '17 at 14:52
If you input more that 4 chars the behavior is undefined due to buffer overflow. — LPs, Mar 17 '17 at 14:53
Step 1: `scanf("%s",msg);` --> `scanf("%4s",msg);` (edit in a `4`) — chux - Reinstate Monica, Mar 17 '17 at 14:53
[Take a look at your code working](http://ideone.com/HW4TQg) — LPs, Mar 17 '17 at 14:53
hey thanks everyone my issue is resolved . it was printing only first charcater because i gave space after the first charcter — hageApple, Mar 17 '17 at 15:09
Possible duplicate of [Reading a string with scanf](http://stackoverflow.com/questions/5406935/reading-a-string-with-scanf) — Rahul Sharma, Mar 17 '17 at 17:51

score 4 · Answer 1 · answered Mar 17 '17 at 15:12

4

The answer is hidden in the comments. The OP states if i give input "i do" it prints only 'i' .

scanf only reads until the first whitespace (which is the i in i do).

If you input 4 characters, 4 characters are printed.

Please edit your question to include this information for future searchers.

answered Mar 17 '17 at 15:12

KevinDTimm

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"scanf only reads until the first whitespace" --> It depends on the format. `"%s"` first reads and discards leading white-space - it does not stop. Then it reads non-whitespace characters until reaching a whitespace. – chux - Reinstate Monica Mar 17 '17 at 16:16
@chux - correct, I answered the question asked by the OP :) – KevinDTimm Mar 17 '17 at 16:22

score 0 · Answer 2 · answered Mar 17 '17 at 17:07

The %s conversion specifier tells scanf to skip over any leading whitespace, then read up to (but not including) the next whitespace character. So when you enter "i do", the %s conversion specifier only reads "i".

If you want to read strings with whitespace, you'll have to use a different method. You can use the %[ conversion specifier like so:

scanf( "%[^\n]", msg );

which will read everything (including leading whitespace characters) until it sees the '\n' newline character and stores it to msg. Unfortunately, this (and the %s specifier above) leave you open to buffer overflow - if your user enters a string longer than msg is sized to hold, scanf will happily write those extra characters to the memory following the msg buffer, leading to all kinds of mayhem. To prevent that, you'd use a field width in the conversion spec, like so:

scanf( "%4[^\n]", msg );

This tells scanf to read no more than 4 characters into msg. Unfortunately, that width has to be hardcoded - you can't pass it as an argument the way you can with printf.

IMO, a better way to go would be to use fgets:

fgets( msg, sizeof msg, stdin );

Why does printf prints only first character of a character array whose input was read through scanf

2 Answers2