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June 6, 2014 15:11
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Minimum Functions for 3 Integers in C
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#include <stdio.h> | |
#include <limits.h> | |
#include <sys/time.h> | |
#define TEST_ITTERATIONS 100 | |
#define TEST_LOOPS 100000 | |
static inline int min3a(int a, int b, int c) | |
{ | |
if (a < b) { | |
if (a < c) { | |
return a; | |
} else { | |
return c; | |
} | |
} else { | |
if (b < c) { | |
return b; | |
} else { | |
return c; | |
} | |
} | |
} | |
static inline int min3b(int a, int b, int c) | |
{ | |
int result = a; | |
if (b < result) { | |
result = b; | |
} | |
if (c < result) { | |
return c; | |
} | |
return result; | |
} | |
static void testA() | |
{ | |
int result = 0; | |
for (int i = 0; i < TEST_LOOPS; ++i) { | |
result = 0; | |
result += min3a(1, 2, 3); | |
result += min3a(1, 3, 2); | |
result += min3a(2, 1, 3); | |
result += min3a(2, 3, 1); | |
result += min3a(3, 1, 2); | |
result += min3a(3, 2, 1); | |
if (result != 6) { | |
printf("Bad min (method A, group 1, loop %d)\n", i); | |
} | |
result = 0; | |
result = min3a(INT_MAX-2, INT_MAX-1, INT_MAX); | |
result = min3a(INT_MAX-2, INT_MAX, INT_MAX-1); | |
result = min3a(INT_MAX-1, INT_MAX-2, INT_MAX); | |
result = min3a(INT_MAX-1, INT_MAX, INT_MAX-2); | |
result = min3a(INT_MAX, INT_MAX-2, INT_MAX-1); | |
result = min3a(INT_MAX, INT_MAX-1, INT_MAX-2); | |
if (result != INT_MAX-2) { | |
printf("Bad min (method A, group 2, loop %d)\n", i); | |
} | |
} | |
} | |
static void testB() | |
{ | |
int result = 0; | |
for (int i = 0; i < TEST_LOOPS; ++i) { | |
result = 0; | |
result += min3b(1, 2, 3); | |
result += min3b(1, 3, 2); | |
result += min3b(2, 1, 3); | |
result += min3b(2, 3, 1); | |
result += min3b(3, 1, 2); | |
result += min3b(3, 2, 1); | |
if (result != 6) { | |
printf("Bad min (method B, group 1, loop %d)\n", i); | |
} | |
result = 0; | |
result = min3b(INT_MAX-2, INT_MAX-1, INT_MAX); | |
result = min3b(INT_MAX-2, INT_MAX, INT_MAX-1); | |
result = min3b(INT_MAX-1, INT_MAX-2, INT_MAX); | |
result = min3b(INT_MAX-1, INT_MAX, INT_MAX-2); | |
result = min3b(INT_MAX, INT_MAX-2, INT_MAX-1); | |
result = min3b(INT_MAX, INT_MAX-1, INT_MAX-2); | |
if (result != INT_MAX-2) { | |
printf("Bad min (method B, group 2, loop %d)\n", i); | |
} | |
} | |
} | |
int main(int argc, char const *argv[]) | |
{ | |
struct timeval stop, start; | |
int totalTime = 0; | |
for (int i = 0; i < TEST_ITTERATIONS; ++i) { | |
gettimeofday(&start, NULL); | |
testA(); | |
gettimeofday(&stop, NULL); | |
totalTime += (stop.tv_usec - start.tv_usec); | |
printf("Method A: %u us \n", stop.tv_usec - start.tv_usec); | |
} | |
printf("A Avg: %u us \n", (totalTime / TEST_ITTERATIONS)); | |
totalTime = 0; | |
for (int i = 0; i < TEST_ITTERATIONS; ++i) { | |
gettimeofday(&start, NULL); | |
testB(); | |
gettimeofday(&stop, NULL); | |
totalTime += (stop.tv_usec - start.tv_usec); | |
printf("Method B: %u us \n", stop.tv_usec - start.tv_usec); | |
} | |
printf("B Avg: %u us \n", (totalTime / TEST_ITTERATIONS)); | |
return 0; | |
} |
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