Return type deduction of lambda expressions of if-else statements

I'm reading C++ primer 5th edition, chapter 10(lambdas expressions), Here is a program that is intended to replace negative values in a vector by their absolute value.

transform(vi.begin(), vi.end(), vi.begin(),
      [](int i) { if (i < 0) return -i; else return i; });

The author says that:

This code won't compile because the lambda infers the return type as void but we returned a value and to fix this, we must use a trailing return type.

But when I compile this code with GNU GCC Compiler on Windows, it works well.

The author also says that:

This version compile because we need not specify the return type, because that type can be inferred from the type of the conditional operator.

transform(vi.begin(), vi.end(), vi.begin(),
          [](int i) { return i < 0 ? -i : i; });

So, my questions are:

• Why with the first version, the lambda infers the return type as void and why does GNU GCC compiler accept this.*(I thought that maybe because of optimizations).?
• Why with the second version, the return type can be inferred from the type of the conditional operator?