83a179abb0
Convenience function to fill a buffer with random bytes. Add some unit tests (only sanity checks, really.)
68 lines
1.9 KiB
C
68 lines
1.9 KiB
C
#include <stdio.h>
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#include <string.h>
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#include "unity.h"
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#include "esp_system.h"
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/* Note: these are just sanity tests, not the same as
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entropy tests
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*/
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TEST_CASE("call esp_random()", "[random]")
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{
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const size_t NUM_RANDOM = 128; /* in most cases this is massive overkill */
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uint32_t zeroes = UINT32_MAX;
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uint32_t ones = 0;
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for (int i = 0; i < NUM_RANDOM - 1; i++) {
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uint32_t r = esp_random();
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ones |= r;
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zeroes &= ~r;
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}
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/* assuming a 'white' random distribution, we can expect
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usually at least one time each bit will be zero and at
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least one time each will be one. Statistically this
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can still fail, just *very* unlikely to. */
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TEST_ASSERT_EQUAL_HEX32(0, zeroes);
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TEST_ASSERT_EQUAL_HEX32(UINT32_MAX, ones);
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}
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TEST_CASE("call esp_fill_random()", "[random]")
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{
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const size_t NUM_BUF = 200;
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const size_t BUF_SZ = 16;
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uint8_t buf[NUM_BUF][BUF_SZ];
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uint8_t zero_buf[BUF_SZ];
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uint8_t one_buf[BUF_SZ];
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bzero(buf, sizeof(buf));
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bzero(one_buf, sizeof(zero_buf));
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memset(zero_buf, 0xFF, sizeof(one_buf));
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for (int i = 0; i < NUM_BUF; i++) {
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esp_fill_random(buf[i], BUF_SZ);
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}
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/* No two 128-bit buffers should be the same
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(again, statistically this could happen but it's very unlikely) */
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for (int i = 0; i < NUM_BUF; i++) {
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for (int j = 0; j < NUM_BUF; j++) {
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if (i != j) {
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TEST_ASSERT_NOT_EQUAL(0, memcmp(buf[i], buf[j], BUF_SZ));
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}
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}
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}
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/* Do the same all bits are zero and one at least once test across the buffers */
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for (int i = 0; i < NUM_BUF; i++) {
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for (int x = 0; x < BUF_SZ; x++) {
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zero_buf[x] &= ~buf[i][x];
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one_buf[x] |= buf[i][x];
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}
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}
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for (int x = 0; x < BUF_SZ; x++) {
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TEST_ASSERT_EQUAL_HEX8(0, zero_buf[x]);
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TEST_ASSERT_EQUAL_HEX8(0xFF, one_buf[x]);
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}
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}
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