How do you format a (Signed) long long integer using printf?

Question

How do you format a (Signed) long long int using printf?

#include <stdio.h>
int main() {
    long long int num = 123456766666666890; //FYI: fits in 29 bits
    int RegularInteger = 4;
    printf("My number is %d bytes wide and its value is %d. A normal number is %d.\n", sizeof(num), num, RegularInteger);
    return 0;
}

Output:

My number is 8 bytes wide and its value is 1917383562. A normal number is 4.

I am not getting desired output.

You may want to read this question and its answers: [ULL suffix on a numeric literal](https://stackoverflow.com/questions/8809292/ull-suffix-on-a-numeric-literal) — CiaPan, Mar 30 '21 at 16:10
My compiler complains: `a.c:4:25: error: integer literal is too large to be represented in any integer type long long int num = 1234567666666666666666890; //FYI: fits in 29 bits` — William Pursell, Mar 30 '21 at 16:10
123456766666666890 == 0x1B69B467248F38A requires more than 29 bits to represent! — William Pursell, Mar 30 '21 at 16:13
@SpirosGkogkas I am using online c compiler. Don't know about the rest https://www.onlinegdb.com/online_c_compiler — , Mar 30 '21 at 16:18
@RetiredNinja Ok, Sorry I am new., I have deleted those comments. — , Mar 30 '21 at 16:20
@SpirosGkogkas Are there any non-C99 compilers that support `long long int` but do not use `%lld` as the corresponding printf/scanf specifier? — Ian Abbott, Mar 30 '21 at 16:28
@IanAbbott avr-gcc for example, many stdlib nano implementations (ARM, RISC-V) etc etc — 0___________, Mar 30 '21 at 16:29
@IanAbbott compiler does not care about formats, implementation in the standard library does — 0___________, Mar 30 '21 at 16:30
@0___________ "compiler does not care about formats, implementation in the standard library does" is not completely true these days. — Ian Abbott, Mar 30 '21 at 16:34
@IanAbbott if you write about compiler printf format check then yes - compiler will checkit, consider OK, no warning, but the implementation will not print it correctly. BTW the same situation is with floats. Embedded libraries do not print floats by default. — 0___________, Mar 30 '21 at 16:40
@0___________ The compiler could also optimize based on the format string, such as replacing `printf("%s\n", string);` with `puts(string);`. — Ian Abbott, Mar 30 '21 at 16:46

score 3 · Answer 1 · answered Mar 30 '21 at 16:55

Enable all warnings, A good compiler will warn about mis-matched specifiers and save time.

Use matching specifiers.

With C99 or later

#include <stdio.h>
int main() {
  long long int num = 123456766666666890; // A 57 bit number
  int RegularInteger = 4;

  // printf("My number is %d bytes wide",  sizeof(num));
  printf("My number is %zu bytes wide",  sizeof(num));

  // printf("and its value is %d.", num);
  printf("and its value is %lld.", num);

  printf(" A normal number is %d.\n", RegularInteger);
  return 0;
}

score 0 · Answer 2 · answered Mar 30 '21 at 16:22

0

#include <stdio.h>
int main() {
    long long int num = 123456766666666890; //FYI: fits in 29 bits
    int RegularInteger = 4;
    printf("My number is %zu bytes wide and its value is %lld. A normal number is %d.\n", sizeof(num), num, RegularInteger);
    return 0;
}

https://godbolt.org/z/eMvjGnr7Y

Source of image

answered Mar 30 '21 at 16:22

0___________

60,014
4
34
74

Any comment silent DV-ter – 0___________ Mar 30 '21 at 16:46
Not my DV (and any DV previously applied seems to have been removed), but this answer would be improved by some explanatory prose. – John Bollinger Mar 30 '21 at 16:58

How do you format a (Signed) long long integer using printf?

2 Answers2