Modifying value of char pointer in c produces segfault?

The standard explicitly lists this as undefined behavior in §J.2: — The program attempts to modify a string literal (6.4.5) If you want to copy it into a local array, do: char a = "abc a is an array on the stack, and you can modify it freely.

The standard explicitly lists this as undefined behavior in §J.2: — The program attempts to modify a string literal (6.4.5) If you want to copy it into a local array, do: char a = "abc"; a is an array on the stack, and you can modify it freely.

Attempting to modify a string literal causes undefined behaviour.

There has to be. – Smoke Manmuscle Mar 4 at 4:49 1 @Smoke, no there is not. You need to copy the literal into an array that's not in read-only memory.

– Carl Norum Mar 4 at 4:50 Thanks that makes sense. – Smoke Manmuscle Mar 4 at 4:53.

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