What is happening in this char* assignment? (comma operator with mixed types)

Question

I'm working on some C code. In one of .c files I can see something like:

char* test = ("someChar", "someChar2", 3);

When I print out this variable, I get "3" on my screen.

What is happening in this part of code? Why do I get 3 as a result of printing out this char*? I am the most curious about this ("someChar", "someChar2", 3) expression.

EDIT(after the issue has been resolved):

What made me scratch my head was also the fact, that there are two chars and one int in this expression. If we use printf("%u", test) we can get this number, but this code definitely doesn't look clean and I believe this is not an elegant way of assigning number to char*.

I think this isn't the code you're actually looking at. If you set a char* to 3 you will most likely get a segmentation fault (or garbage output) when trying to print it. Maybe it's "3" ? — cleblanc, Jun 06 '18 at 13:56
Something like is not a very precise definition. What is exactly the line and how is used `test` later? — Serge Ballesta, Jun 06 '18 at 14:01
Thank you @FredLarson - it indeed looks like the comma operator — Paradowski, Jun 06 '18 at 14:02
Possible duplicate of [What does the comma operator , do?](https://stackoverflow.com/questions/52550/what-does-the-comma-operator-do) — Fred Larson, Jun 06 '18 at 14:02
Hellofa way to generate a pointer with address `3` via the *comma operator* and a parenthetical grouping. Attempting to access address `3` will come with a sudden surprise.. — David C. Rankin, Jun 06 '18 at 19:33

Achal · Accepted Answer · 2018-06-06T14:09:25.733

3

Its because of comma operator & manual page of operator says when in an expression if multiple comma are there then solve from Left to Right but it considers right most argument.

In the statement

char* test = ("someChar", "someChar2", 3);

test get assigned with right most argument that is 3. And now it looks like

char *test = 3;

since test is char pointer & it should initialize with valid address and 3 is not the valid address. So if you are just printing test like printf("%d\n",test); that doesn't cause any error but that causes undefined behavior. And if you are going to dereference it like *test then your program get crashed(Seg. fault), this is one of possible scenario you should keep in mind while dealing with pointers.

edited Jun 06 '18 at 14:09

answered Jun 06 '18 at 14:02

Achal

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printing a char* with `%d` is still undefined behavior – cleblanc Jun 06 '18 at 14:04
yup @cleblanc agree. I thought of writing it but forgot to put it. modified. – Achal Jun 06 '18 at 14:07
Thank you for explanation. To make things clear: the int value enclosed in my post in this assignment is not there by mistake. However I don't know why the original programmer put there this integer. – Paradowski Jun 06 '18 at 14:23
Welcome !! May be original programmer wanted to know whether one understood comma operator correctly or not. But you should be aware of all possible scenario specially with pointers. – Achal Jun 06 '18 at 14:27

What is happening in this char* assignment? (comma operator with mixed types)

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