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struct S {
  // Assume these have side effects
  int f(char);
  double f(double);
};
 
void f() {
  S s;
  (void)typeid(s.f(1.0));
}

Some expressions involve operands that are unevaluated. According to the C++ Standard, [expr], paragraph 8 [ISO/IEC 14882-2014]:

In some contexts, unevaluated operands appear. An unevaluated operand is not evaluated. An unevaluated operand is considered a full-expression. [Note: In an unevaluated operand, a non-static class member may be named (5.1) and naming of objects or functions does not, by itself, require that a definition be provided. — end note]

The following expressions do not evaluate their operands: sizeof(), typeid(), noexcept(), decltype(), and declval().

Because an unevaluated operand in an expression is not evaluated, no side effects from that operand will be triggered. Reliance on those side effects will result in unexpected behavior. Do not rely on side effected effects in unevaluated operands.The following expressions do not evaluate their operands: sizeof(), typeid(), noexcept(), decltype(), and declval()

Note that unevaluated expression operands are used in situations were the declaration of an object is required, but the definition of the object is not. For instance, in the example below, the function f() is overloaded, which relies on the unevaluated expression operand to select the desired overload, which is then used to determine the result of the sizeof() expression:

int f(int);
double f(double);
 
size_t size = sizeof(f(0));

Noncompliant Code Example (sizeof)

...

In this noncompliant code example, the call to g() is expression i++ is not evaluated within the decltype specifier:

#include <iostream>

static int glob = 100;

int gvoid f() {
 return ++glob; }

void f() {int i = 0;
  decltype(g()i++) h = 12;
  std::cout << globi;
}

Consequently, the value of glob i remains 14. The function call syntax is used within decltype to distinguish between the return type of g() and the function type of g, but the call is never evaluated.0.

Compliant Solution (decltype)

In this compliant solution, g() i is called incremented outside of the decltype specifier, so that it is evaluated as desired:

#include <iostream>

static int glob = 100;

int gvoid f() {
 return ++glob; }

void f() {int i = 0;
  decltype(g()i) h = 12;
  g()++i;
  std::cout << globi;
}

Risk Assessment

If expressions that appear to produce side effects are an unevaluated operand, the results may be different than expected. Depending on how this result is used, it can lead to unintended program behavior.

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