Differing output with printf using %d and %i

Question

I found questions about %i and %d on here, but all of them seemed to claim that they were the same in printf.
Compiler: Apple clang version 12.0.5 (clang-1205.0.22.9)
Note: 15 is 017 in octal and 0xf in hexidecimal.

I was previously under the impression that %d and %i only differed when used during scanf and not printf. Is this implementation defined behavior? Additionally I thought the number would have been converted and stored prior to the call to scanf.

Output:
scanf using %i: 017 017 0xf 0xf //user input
%i: 15, %d: 15, %i: 15, %d: 15

scanf using %d: 010 010 0xf 0xf //user input
%i: 10, %d: 10, %i: 0, %d: 15 //They all make sense to me except these last two

Code:

int a, b, c, d;     
                        
printf("scanf using %%i: ");    
scanf("%i %i %i %i", &a, &b, &c, &d);    
printf("%%i: %i, %%d: %d, %%i: %i, %%d: %d\n\n",a,b,c,d);    
                        
printf("scanf using %%d: ");    
scanf("%d %d %d %d", &a, &b, &c, &d);    
printf("%%i: %i, %%d: %d, %%i: %i, %%d: %d\n\n",a,b,c,d);

Can you clarify what you mean "scanf failed", code compiles cleanly. — Clay Smith, Commented Aug 5, 2022 at 6:28
Check what scanf returns. Also read more about the differences between %i and %d in scanf (they do not work the same). — Some programmer dude, Commented Aug 5, 2022 at 6:29
You are right scanf returns 4 on the first one (as expected), but only 3 (not as expected, by me at least) on the last one. Can you explain why that might be? — Clay Smith, Commented Aug 5, 2022 at 6:32
As explained in the (first) answer, 0xf is interpreted by %d as 0, and the x is not valid as part of a decimal integer, so scanf() fails to read a fourth value and reports that it read just 3 values. It does not modify the fourth variable, so it continues to store what was in it before the call. — Jonathan Leffler, Commented Aug 5, 2022 at 6:36

2 revs, 2 users 94% · Accepted Answer · 2022-08-05 06:33:34Z

1

scanf("%d %d %d %d", &a, &b, &c, &d); scans 010 010 0 from the input and assigns: 10 to a, 10 to b and 0 to c.

Then xf 0xf is left in the input. Because character 'x' is invalid for a number %d for &d, so scanf fails the scanning and d is left not assigned and scanf() returns 3 rather than 4 to let you know it failed to read 4 values.

So c has the value of 0, and d has the previous value of 15.

Is this implementation defined behavior?

No, it is defined.

edited Aug 5, 2022 at 6:33

community wiki

2 revs, 2 users 94%
KamilCuk

2

Thank you! This makes sense to me now. Thank you for clarifying, I will mark this as solved then when the 5 minutes is up and they allow me to! Thanks again for the explanation.
– Clay Smith
Commented Aug 5, 2022 at 6:35

Add a comment |

feiniks · Accepted Answer · 2022-08-05 06:45:31Z

Each pointer argument of scanf() must be of a type that is appropriate for the value returned by the corresponding conversion specification. If the number of conversion specifications in format exceeds the number of pointer arguments, the results are undefined. If the number of pointer arguments exceeds the number of conversion specifications, then the excess pointer arguments are evaluated, but are otherwise ignored.

When the format is '%d', your input should also be a decimal number.

scanf using %d: 10 20 33 44 //user input

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