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authorAbseil Team <absl-team@google.com>2022-11-16 17:17:58 (GMT)
committerCopybara-Service <copybara-worker@google.com>2022-11-16 17:18:37 (GMT)
commit9c332145b71c36a5bad9688312c79184f98601ff (patch)
treef465e472334c8dd173ff3cfb76823ed1f18a7f36 /googletest/include
parent4408a0288b24fbe9156a9b14bbecd148ea64f40f (diff)
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When printing floating-point numbers, print full precision by default.
To make debug output readable, we still use the faster 6-digit precision sometimes, but only if it will round-trip. This way, when a test fails due to a very small difference in floating-point numbers, users will have enough digits to see the difference. PiperOrigin-RevId: 488958311 Change-Id: Ibcac43f48a97006d89217530c69386cc4fa2735c
Diffstat (limited to 'googletest/include')
-rw-r--r--googletest/include/gtest/gtest-printers.h75
1 files changed, 75 insertions, 0 deletions
diff --git a/googletest/include/gtest/gtest-printers.h b/googletest/include/gtest/gtest-printers.h
index 89215e6..eeaef04 100644
--- a/googletest/include/gtest/gtest-printers.h
+++ b/googletest/include/gtest/gtest-printers.h
@@ -484,6 +484,81 @@ GTEST_API_ void PrintTo(__uint128_t v, ::std::ostream* os);
GTEST_API_ void PrintTo(__int128_t v, ::std::ostream* os);
#endif // __SIZEOF_INT128__
+// The default resolution used to print floating-point values uses only
+// 6 digits, which can be confusing if a test compares two values whose
+// difference lies in the 7th digit. So we'd like to print out numbers
+// in full precision.
+// However if the value is something simple like 1.1, full will print a
+// long string like 1.100000001 due to floating-point numbers not using
+// a base of 10. This routiune returns an appropriate resolution for a
+// given floating-point number, that is, 6 if it will be accurate, or a
+// max_digits10 value (full precision) if it won't, for values between
+// 0.0001 and one million.
+// It does this by computing what those digits would be (by multiplying
+// by an appropriate power of 10), then dividing by that power again to
+// see if gets the original value back.
+// A similar algorithm applies for values larger than one million; note
+// that for those values, we must divide to get a six-digit number, and
+// then multiply to possibly get the original value again.
+template <typename FloatType>
+int AppropriateResolution(FloatType val) {
+ int full = std::numeric_limits<FloatType>::max_digits10;
+ if (val < 0) val = -val;
+
+ if (val < 1000000) {
+ FloatType mulfor6 = 1e10;
+ if (val >= 100000.0) { // 100,000 to 999,999
+ mulfor6 = 1.0;
+ } else if (val >= 10000.0) {
+ mulfor6 = 1e1;
+ } else if (val >= 1000.0) {
+ mulfor6 = 1e2;
+ } else if (val >= 100.0) {
+ mulfor6 = 1e3;
+ } else if (val >= 10.0) {
+ mulfor6 = 1e4;
+ } else if (val >= 1.0) {
+ mulfor6 = 1e5;
+ } else if (val >= 0.1) {
+ mulfor6 = 1e6;
+ } else if (val >= 0.01) {
+ mulfor6 = 1e7;
+ } else if (val >= 0.001) {
+ mulfor6 = 1e8;
+ } else if (val >= 0.0001) {
+ mulfor6 = 1e9;
+ }
+ if (static_cast<int32_t>(val * mulfor6 + 0.5) / mulfor6 == val) return 6;
+ } else if (val < 1e10) {
+ FloatType divfor6 = 1.0;
+ if (val >= 1e9) { // 1,000,000,000 to 9,999,999,999
+ divfor6 = 10000;
+ } else if (val >= 1e8) { // 100,000,000 to 999,999,999
+ divfor6 = 1000;
+ } else if (val >= 1e7) { // 10,000,000 to 99,999,999
+ divfor6 = 100;
+ } else if (val >= 1e6) { // 1,000,000 to 9,999,999
+ divfor6 = 10;
+ }
+ if (static_cast<int32_t>(val / divfor6 + 0.5) * divfor6 == val) return 6;
+ }
+ return full;
+}
+
+inline void PrintTo(float f, ::std::ostream* os) {
+ auto old_precision = os->precision();
+ os->precision(AppropriateResolution(f));
+ *os << f;
+ os->precision(old_precision);
+}
+
+inline void PrintTo(double d, ::std::ostream* os) {
+ auto old_precision = os->precision();
+ os->precision(AppropriateResolution(d));
+ *os << d;
+ os->precision(old_precision);
+}
+
// Overloads for C strings.
GTEST_API_ void PrintTo(const char* s, ::std::ostream* os);
inline void PrintTo(char* s, ::std::ostream* os) {