C++ pointer to local string literal

Question

I came across this code bellow today and I am not totally sure what it will do:

pLogFileCriteria->taskToLogFor[0][0] = *"*";

It's setting the value to a pointer to a local string that's on the stack? So when it exits this function, does that memory still exist? Seems fish to me.

Solomon Slow · Accepted Answer · 2017-08-25 20:44:31Z

4

The type of *"*" is const char. Your code snippet is assigning a char value to the location computed by the left-hand side.

This would do the same thing:

pLogFileCriteria->taskToLogFor[0][0] = '*';

edited Aug 25, 2017 at 20:44

answered Aug 25, 2017 at 20:31

Solomon Slow

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2 Comments

Tyrel Van Niekerk Over a year ago

You are probably correct. It just seems like the people who wrote this code wanted to make it as complicated as possible.

Barmar Over a year ago

@TyrelVanNiekerk If that's what they were going for, they didn't try very hard. It could be far more complicated.

The Techel · Accepted Answer · 2017-08-25 20:31:47Z

1

String literals have static duration. They exist for the life of the program, thus obtaining a pointer to it's first character will remain valid after the function. But note the indirect operator. It will cause the pointer to the first character to be dereferenced, yielding that character.

answered Aug 25, 2017 at 20:31

The Techel

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1 Comment

Tyrel Van Niekerk Over a year ago

So it's really just setting that character to '*' as james said in his comment.

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