div_t div( int x, int y );
(1)ldiv_t ldiv( long x, long y );
(2)lldiv_t lldiv( long long x, long long y );
(3) (since C99) (4) (since C99)Computes both the quotient and the remainder of the division of the numerator x
by the denominator y
.
Computes quotient and remainder simultaneously. The quotient is the algebraic quotient with any fractional part discarded (truncated towards zero). The remainder is such that quot * y + rem == x.
(until C99)Computes the quotient (the result of the expression x / y) and remainder (the result of the expression x % y) simultaneously.
(since C99) [edit] Parameters [edit] Return valueIf both the remainder and the quotient can be represented as objects of the corresponding type (int, long, long long, intmax_t, respectively), returns both as an object of type div_t
, ldiv_t
, lldiv_t
, imaxdiv_t
defined as follows:
struct div_t { int quot; int rem; };
or
struct div_t { int rem; int quot; };ldiv_t
struct ldiv_t { long quot; long rem; };
or
struct ldiv_t { long rem; long quot; };lldiv_t
struct lldiv_t { long long quot; long long rem; };
or
struct lldiv_t { long long rem; long long quot; };
If either the remainder or the quotient cannot be represented, the behavior is undefined.
[edit] NotesUntil C99, the rounding direction of the quotient and the sign of the remainder in the built-in division and remainder operators was implementation-defined if either of the operands was negative, but it was well-defined in div
and ldiv
.
On many platforms, a single CPU instruction obtains both the quotient and the remainder, and this function may leverage that, although compilers are generally able to merge nearby / and % where suitable.
[edit] Example#include <assert.h> #include <limits.h> #include <math.h> #include <stdio.h> #include <stdlib.h> void reverse(char* first, char* last) { for (--last; first < last; ++first, --last) { char c = *last; *last = *first; *first = c; } } // returns empty buffer in case of buffer overflow char* itoa(int n, int base, char* buf, size_t buf_size) { assert(2 <= base && base <= 16 && buf && buf_size); div_t dv = {.quot = n}; char* p = buf; do { if (!--buf_size) return (*buf = '\0'), buf; dv = div(dv.quot, base); *p++ = "0123456789abcdef"[abs(dv.rem)]; } while(dv.quot); if (n < 0) *p++ = '-'; *p = '\0'; reverse(buf, p); return buf; } int main(void) { char buf[16]; printf("%s\n", itoa(0, 2, buf, sizeof buf)); printf("%s\n", itoa(007, 3, buf, sizeof buf)); printf("%s\n", itoa(12346, 10, buf, sizeof buf)); printf("%s\n", itoa(-12346, 10, buf, sizeof buf)); printf("%s\n", itoa(-42, 2, buf, sizeof buf)); printf("%s\n", itoa(INT_MAX, 16, buf, sizeof buf)); printf("%s\n", itoa(INT_MIN, 16, buf, sizeof buf)); }
Possible output:
0 21 12346 -12346 -101010 7fffffff -80000000[edit] References
RetroSearch is an open source project built by @garambo | Open a GitHub Issue
Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo
HTML:
3.2
| Encoding:
UTF-8
| Version:
0.7.4