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V1063. The modulo by 1 operation is mea…
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# V1063. The modulo by 1 operation is meaningless. The result will always be zero.

Aug 12 2020

The analyzer has detected a strange expression with a modulo by 1 operation. Such an expression will always evaluate to 0.

A common pattern of this error is checking if no remainder is left after dividing a value by another value. To do this, you use the modulo operation and compare the result with 0 or 1. Making a typo at this point is easy because since you anticipate the value 1, you may accidentally divide by 1 too. For example:

``````if (x % 1 == 1)
{
....
}``````

A modulo by 1 operation was applied to the 'x' variable, which will result in the 'x % 1' expression always evaluating to 0 no matter the value of 'x'. Therefore, the condition will always be false. The programmer must have intended to use the modulo by '2' operation:

``````if (x % 2 == 1)
{
....
}``````

The following example is taken from a real application (stickies):

``````void init (....)
{
srand(GetTickCount() + rand());

updateFreq1 = (rand() % 1) + 1;
updateFreq2 = (rand() % 1) + 1;
updateFreq3 = (rand() % 1) + 1;
updateFreq4 = (rand() % 1) + 1;

waveFreq1 = (rand() % 15);
waveFreq2 = (rand() % 3);
waveFreq3 = (rand() % 16);
waveFreq4 = (rand() % 4);
// ....
}``````

The variables 'updateFreq1', 'updateFreq2', 'updateFreq3', and 'updateFreq4' will always be initialized to the value 1. Each of these variables was probably meant to be initialized to some pseudorandom value, which most likely falls within the range [1..2]. In that case, the correct version should look like this:

``````updateFreq1 = (rand() % 2) + 1;
updateFreq2 = (rand() % 2) + 1;
updateFreq3 = (rand() % 2) + 1;
updateFreq4 = (rand() % 2) + 1;``````

This diagnostic is classified as:

 You can look at examples of errors detected by the V1063 diagnostic.