printf of int value returns only first character

Question

When I run the code below i get this output:

Podaj ilosc liczb: 12

Podaj liczbe nr. 0 : 53

Podaj liczbe nr. 1 : 24

Podaj liczbe nr. 2 : 53

Podaj liczbe nr. 3 : 24

Podaj liczbe nr. 4 : 66

Podaj liczbe nr. 5 : 99

Podaj liczbe nr. 6 : 3

Podaj liczbe nr. 7 : 0

Podaj liczbe nr. 8 : 5

Podaj liczbe nr. 9 : 2

Podaj liczbe nr. 10 : 5

Podaj liczbe nr. 11 : 2 Twoje liczby w odwrotnej kolejnosci: 2 5 2 5 0 3 9 6 2 5 2 5

#include <stdio.h>

int main(int argc, char **argv)
{

char tablo[100];
int k,i;
printf("Podaj ilosc liczb: ");
scanf("%d", &k);fflush(stdin);
for(i=0; i<k; i++){
                 printf("\nPodaj liczbe nr. %d : ", i);
                 scanf("%c", &tablo[i]);fflush(stdin);
                 }
printf("Twoje liczby w odwrotnej kolejnosci: \n");
for(i=k-1;i>=0;i--){
                    printf("%3c",tablo[i]);
                    }
                    printf("\n \n");

return 0;
}

Why? How to fix it? I just want my numbers in the reverse sequence.

use `puts` to just print a string without any formatting, calling `printf` without any arguments other than the formatting string for that purpose is undefined behavior. — Mgetz, Nov 12 '14 at 18:45
@Mgetz Disagree with "calling printf without any arguments other than the formatting string for that purpose is undefined behavior". Calling `printf()` with incorrect or insufficient agreements is UB. Simple calling `printf("Hello")` is not UB. Further `printf("Podaj ilosc liczb: ")` is not the same as `puts("Podaj ilosc liczb: ")` as the latter adds `\n`. Better to suggest `fputs()` — chux - Reinstate Monica, Nov 12 '14 at 18:55
possible duplicate of [using scanf in pointers to integer array](http://stackoverflow.com/questions/18378346/using-scanf-in-pointers-to-integer-array) — Mihai Maruseac, Nov 29 '14 at 02:34

score 1 · Answer 1 · answered Nov 12 '14 at 18:26

You're reading and writing individual characters, not integers. Try running your program with, say, x or * as input; it will just print the characters you gave it without trying to interpret them as decimal integers.

To read integers (interpreting the sequence "123" as the decimal number 123), use scanf("%d", ...) to read into an int object, and printf("%d", ...) to print the values -- and define tablo as an array of int rather than of char. (And don't forget to check the value returned by scanf, so you can tell whether it succeeded or failed.)

And don't use fflush(stdin). The behavior of fflush on input streams is undefined, and you don't really need it anyway. Later you might want a way to discard invalid input, but fflush isn't the answer.

Alternately, he could simply change the conversion specifier to `%hhd` if his values will all fit into the `char` data type. — John Bode, Nov 12 '14 at 18:34
@JohnBode only if `char` is signed by default. Otherwise, it'd have to be `%hhu` — Drew McGowen, Nov 13 '14 at 15:52

score 1 · Answer 2 · answered Nov 12 '14 at 18:44

The %c conversion specifier tells scanf to read a single character from the input stream; when you type "53", only the character '5' will be read. If you want to read numerical values and store them in a char data type, you need to use the %hhd conversion specifier:

scanf( "%hhd", &tablo[i] );

Note that if you want to store any value greater than 128, you'll need to use a wider data type such as int and the appropriate conversion specifier (%d for int, %hd for short, %ld for long, etc).

Don't use fflush on input streams like stdin; the behavior isn't defined, and it may not do what you intend.

score 0 · Accepted Answer · answered Nov 12 '14 at 18:24

0

You're only scanning for single characters, not actual numbers. Instead, tablo should be an array of int (not char), and instead of using %c for scanf and printf, you should use %d.

answered Nov 12 '14 at 18:24

Drew McGowen

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printf of int value returns only first character

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