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);
CodePudding user response:
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.
CodePudding user response:
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