Created
October 23, 2017 19:44
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剰余計算覚え書き(C言語)
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#include<stdio.h> | |
#include<stdlib.h> | |
int main() | |
{ | |
int a[4] = {3, -3, 3, -3}; | |
int b[4] = {8, 8, -8, -8}; | |
int i = 0; | |
div_t r, rx; | |
printf("a%%b \n"); | |
for (i = 0; i < 4; i++) | |
printf(" %2d %% %2d = %2d \n", a[i], b[i], a[i] % b[i]); | |
printf("(a+b)%%b \n"); | |
for (i = 0; i < 4; i++) | |
printf(" %2d %% %2d = %2d \n", a[i], b[i], (a[i] + b[i]) % b[i]); | |
printf("a&(b-1) \n"); | |
for (i = 0; i < 4; i++) | |
printf(" %2d & (%2d-1) = %2d \n", a[i], b[i], a[i] & (b[i] - 1)); | |
printf("(a%%b+b)%%b \n"); | |
for (i = 0; i < 4; i++) | |
printf(" %2d %% %2d = %2d \n", a[i], b[i], (a[i] % b[i] + b[i]) % b[i]); | |
printf("r=div(a,b), r.quot, r.rem \n"); | |
for (i = 0; i < 4; i++) { | |
r = div(a[i], b[i]); | |
printf(" %2d %% %2d = %2d ... %2d \n", a[i], b[i], r.quot, r.rem); | |
} | |
printf("r=div(a,b), rx=div(r.rem+b, b) , rx.quot, rx.rem\n"); | |
for (i = 0; i < 4; i++) { | |
r = div(a[i], b[i]); | |
rx = div(r.rem + b[i], b[i]); | |
printf(" %2d %% %2d = %2d ... %2d \n", a[i], b[i], rx.quot, rx.rem); | |
} | |
} |
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DSPのために配列を使ったリングバッファの扱いを書き直しているときに罠に嵌ったのでメモ。いわゆる「負数の剰余」問題である。