[glsl-in] True and false branch of conditional expression operands are always evaluated

[DDoS]

The expression:

localSums[left] = globalLeft < globalSumBuffer.count ? globalSumBuffer.sums[globalLeft] : 0;

compiles to:

let _e36 = globalLeft;
let _e38 = globalSumBuffer.count;
let _e40 = globalLeft;
let _e43 = globalSumBuffer.sums[_e40];
localSums[_e34] = select(u32(0), _e43, (_e36 < _e38));

which evaluates both the true and false branches before evaluation the condition. This is wrong according to the GLSL spec:

The ternary selection operator (?:). It operates on three expressions (exp1 ? exp2 : exp3). This
operator evaluates the first expression, which must result in a scalar Boolean. If the result is true, it
selects to evaluate the second expression, otherwise it selects to evaluate the third expression. Only
one of the second and third expressions is evaluated.
GLSL 4.4 Spec page 109

In the example above, if the condition is false, then the true branch is an out-of-bound access, which cannot be evaluated.