Assign to a string literal an array

Question

It is undefined behavior to do this?

char *a = "hello";
char b[] = "abcd";
a = b;

My compiler throws no warning, with the maximum warning level.

merlin2011 · Accepted Answer · 2016-12-27 22:36:20Z

10

There is no UB here. You are simply re-assigning a pointer to point at the address of the start of the array instead.

Note that you are not actually modifying the value that a is pointing at, only a itself, and a is a normal char *.

answered Dec 27, 2016 at 22:36

merlin2011

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2 Comments

cheroky Over a year ago

What happen with "hello" is lost?

merlin2011 Over a year ago

@cheroky, If you had allocated that memory, it would be a memory leak because you could never free it. However, since it is a string literal whose memory was allocated by the compiler (likely in a read-only region), you could never free it anyways, so not having a pointer to it is irrelevant.

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