The output of the code will be ____. int i=5; printf(“%d %d %d”, i, i++, ++i);
Question
The output of the code will be ____. int i=5; printf(“%d %d %d”, i, i++, ++i);
Solution
The output of this code is undefined and depends on the compiler. This is because the order of evaluation of arguments in a function call is unspecified according to the C and C++ standards.
In the expression printf(“%d %d %d”, i, i++, ++i);, the variables i, i++, and ++i are all arguments to the function printf. The order in which these arguments are evaluated is not specified, so different compilers may evaluate them in different orders.
Therefore, the output could be any of 7 5 7, 7 6 7, 6 6 7, etc. depending on the compiler. It's generally a good idea to avoid writing code with such ambiguities.
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