C++ multiple type assignment

Question

I not very familiar with C++, have some issues with the multiple assignments

int a = 0, b = 0;

works

int i; double d; char c; c = i = d = 42;

also works but why this does not works.

int a = 4, float b = 4.5, bool ab = TRUE;

It doesn't work because those are the rules of C++. Differing types cannot be declared in a single line like that. — PaulMcKenzie
– PaulMcKenzie, Commented May 15, 2021 at 22:21
Because the language was designed that way: you can only declare multiple variables with the same base type. Also note that your bool should be true, not TRUE. — Quentin
– Quentin, Commented May 15, 2021 at 22:21
But int a = 4; float b = 4.5; bool ab = true; // in one line works. — user15388024
– user15388024, Commented May 15, 2021 at 22:23
@CEPB "I think c = i = d = 42; is UB" no, this is well defined. "What if you do c = i = d = 55555" yes, that's UB if it causes a signed overflow. For unsigned wrap around it's still well defined. char can be unsigned or signed. It implementation defined. — user15388024
– user15388024, Commented May 15, 2021 at 22:27

tadman · Accepted Answer · 2021-05-15 23:19:56Z

This is to do with the allowed syntax. The format is, in very high-level terms, roughly:

type varName [ = expr ][, varName [ = expr ] ]?;

Where there is no allowance at all for another type to be introduced mid-stream.

In practice declaring multiple variables per line can lead to a lot of ambiguity and confusion.

Quick quiz: What is the initial value of a? What type is b?

int* a, b = 0;

If you guessed a was a NULL pointer and b was int* you'd be wrong. This is why it's often preferable to spell it out long-form so there's absolute clarity:

int* a; int b = 0;

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