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V6130. Integer overflow in arithmetic expression.
The analyzer has detected an arithmetic operation that may result in an integer overflow.
The example:
private static long parseHumanLong(String str) {
char tail = str.charAt(str.length() - 1);
long base = 1;
switch (tail) {
case 't':
base *= 1000 * 1000 * 1000 * 1000;
break;
case 'b':
base *= 1000 * 1000 * 1000;
break;
case 'm':
base *= 1000 * 1000;
break;
case 'k':
base *= 1000;
break;
default:
}
if (base != 1) {
str = str.substring(0, str.length() - 1);
}
return Long.parseLong(str) * base;
}This method reads numbers and converts their suffixes to:
t– trillion;b– billion;m– million;k– thousand.
If the 1m string is passed to the method, it is expected to be converted to the long variable with the 1_000_000 value.
When calculating trillions in the 1000 * 1000 * 1000 * 1000 expression, the multiplication is performed within the int range, but the resulting number exceeds the maximum value for the int type. This results in the overflow and incorrect result.
To evaluate the expression correctly, explicitly specify the Long type for it.
base *= 1000L * 1000 * 1000 * 1000;This diagnostic is classified as:
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