foo(a = b+c); //new value of a(after the call) = b+c //but sizeof(a = b+c); //new value of a = old value of a Why isn't the the result of the assignment statement reflected in the stack of the function( which contains the above code) in the latter case?
sizeof is an operator not a function. Operand of sizeof is not evaluated except when it is a variable length array.
C11: 6.5.3.4 p(2):
The
sizeofoperator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.
sizeof does not evaluate its operand.
size_t x = sizeof(i++); // i is not incremented Except when variable length arrays are involved:
(C99, 6.5.3.4p2) "If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant."
size_t y = sizeof(int [j++]); // j is incremented; (C99, 6.7.5.2p4) "Where a size expression is part of the operand of a sizeof operator and changing the value of the size expression would not affect the result of the operator, it is unspecified whether or not the size expression is evaluated."
size_t z = sizeof (*(int (*)[k++]) 0); // k may or may not be incremented // gcc increments k 
++k, would it be mandatory to increment k? btw.. does this expression invoke UB due to NULL dereference?
sizeof.