Right way to show numbers in binary/hex/octal in your opinion?

So I have this function that converts a number to a string and it can return it in any base you want. However, for negative numbers, the result may not be what you expected for bases other than decimal if you have used "printf" or "wirtef". Let's say that we want to convert then number -10 from decimal to hex. There are two possible results (check here: Decimal to Hexadecimal Converter).

The first one is negate the number and convert it to hex and then add a "-" in front of the number. So the result will be: -a (which is what my function does)

The second one is what "printf" and "writef" does which is using 2's complement. So in this case, we first convert the number to binary using 2's complement and then we convert this number to hex. In this example, -10 to binary is 1111111111110110 which is fff6 in hex. However, for some reason "printf" and "writef" return fffffff6 so go figure....

Here are the advantages of these two methods in my humble opinion:

First method:
1. Is is what my function does and I would prefer not to change it obviously. Also implementing the other behavior, would need a lot of work (and of course I will have to figure it out, don't know how to do it) and it will make the function much slower.
2. It is easier on they eyes as it makes it more obvious to understand if the number is signed or not and so what the equivalent is in other systems.

Second method:
1. It is probably what people would expect and what makes scientifically more sense as decimal was supposed to be the only base that will make sense for humans to read hence be the only base that has the "-" character.

Anyway, I don't even know why it will be practical useful to print a number in another system so, what are your thoughts?

Quote:

Originally Posted by rempas

In this example, -10 to binary is 1111111111110110 which is fff6 in hex. However, for some reason "printf" and "writef" return fffffff6 so go figure....

That's because int is presumably 4 bytes and thus 32 bits on your system. The online calculator you used set it as 16 bits instead.

That would be a general issue if you choose to go with option 2: the representation of the negative integer necessarily depends on how many bits you intend to represent. (That's kind of true for positive integers too, but I guess it is more natural for humans familiar with the Hindu-Arabic numeral system to omit leading zeroes.)

It's kind of a silly question. Do what you want. The %o and %x formats for printf are defined to work only with unsigned integers so they interpret the representation of the signed value as an unsigned value.

BTW, the correct answer for a negative signed integer 2's complement representation depends on the number of bits of the variable. For 16 bits, -10 is fff6, but for 32 bits (which is what your system uses) it is fffffff6.

BTW2, there is no such thing as writef.

Code:

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> void print_unsigned(unsigned n, unsigned b) // b in [2, 36] { const char *a = "0123456789abcdefghijklmnopqrstuvwxyz"; if (n >= b) print_unsigned(n / b, b); putchar(a[n % b]); } void print_signed(int n, int b) // b in [2, 36] { if (n < 0) { putchar('-'); n = -n; } print_unsigned(n, b); } int main() { char line[256]; while (printf(">>> "), fgets(line, sizeof line, stdin) != NULL) { int n = 0; if (strcmp(line, "min\n") == 0) n = INT_MIN; else if (strcmp(line, "min+1\n") == 0) n = INT_MIN + 1; else if (strcmp(line, "max\n") == 0) n = INT_MAX; else if (strcmp(line, "-max\n") == 0) n = -INT_MAX; else n = atoi(line); printf("d: %d\n", n); for (int b = 2; b <= 36; ++b) { printf("%2d: ", b); print_signed(n, b); putchar('\n'); } } putchar('\n'); return 0; }

Quote:

Originally Posted by john.c

It's kind of a silly question. Do what you want. The %o and %x formats for printf are defined to work only with unsigned integers so they interpret the representation of the signed value as an unsigned value.

BTW, the correct answer for a negative signed integer 2's complement representation depends on the number of bits of the variable. For 16 bits, -10 is fff6, but for 32 bits (which is what your system uses) it is fffffff6.

BTW2, there is no such thing as writef.

Code:

#include <stdio.h> #include <stdlib.h> #include <string.h> #include <limits.h> void print_unsigned(unsigned n, unsigned b) // b in [2, 36] { const char *a = "0123456789abcdefghijklmnopqrstuvwxyz"; if (n >= b) print_unsigned(n / b, b); putchar(a[n % b]); } void print_signed(int n, int b) // b in [2, 36] { if (n < 0) { putchar('-'); n = -n; } print_unsigned(n, b); } int main() { char line[256]; while (printf(">>> "), fgets(line, sizeof line, stdin) != NULL) { int n = 0; if (strcmp(line, "min\n") == 0) n = INT_MIN; else if (strcmp(line, "min+1\n") == 0) n = INT_MIN + 1; else if (strcmp(line, "max\n") == 0) n = INT_MAX; else if (strcmp(line, "-max\n") == 0) n = -INT_MAX; else n = atoi(line); printf("d: %d\n", n); for (int b = 2; b <= 36; ++b) { printf("%2d: ", b); print_signed(n, b); putchar('\n'); } } putchar('\n'); return 0; }

Thanks! I will use the first way in that case as it is both easier (both on code and on the eyes). Also, I posted this message in the D language forums and "writef" is from the standard library of Dlang. So yeah I forgot to remove it here... ;)

Quote:

Originally Posted by laserlight

That's because int is presumably 4 bytes and thus 32 bits on your system. The online calculator you used set it as 16 bits instead.

That would be a general issue if you choose to go with option 2: the representation of the negative integer necessarily depends on how many bits you intend to represent. (That's kind of true for positive integers too, but I guess it is more natural for humans familiar with the Hindu-Arabic numeral system to omit leading zeroes.)

Thanks a lot! In this case I think it makes more sense to use the first one then and it is also ready so that's awesome!!!