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Asked: 2026-05-24T13:04:57+00:00

As far as I am aware, the linker tries to merge two string literals

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As far as I am aware, the linker tries to merge two string literals into one single literal, if they are both the same, e.g.:

file1.c

char const* firstString = "foo";

file2.c

char const* secondString = "foo";

Would result in only one occurence of foo\0 in the respective memory section (saving 4 Bytes). This is especially important for embedded applications (how does avr-gcc vs. gcc behave).

But I was wondering if I can actually count on this to happen, and rely, that if two strings are equal, also their pointers are equal (provided that in the whole program, you only pass string literals around and no runtime generated strings exist — which is a reasonable assumption in my case). Obviously, I want to speed up speed comparisons with this, and allow a commonly used function to receive a string literal like so:

void lock(char const*);
void unlock(char const*);

lock("test");
dosmth();
unlock("test");

In essence, I want to avoid having a huge enum and huge switches inside the lock/unlock functions.

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1 Answer

  1. Editorial Team

    The linker doesn’t have to merge anything. It just has to map a declared type to a defined type. That involves finding the defined type and filling out the address offsets to jump to the right item.

    What you are talking about would be an optimizing linker. Many linkers don’t optimize at all, and those that do aren’t held to an optimizing standard, so you’ll never be able to generalize beyond the observed findings for the linker you discover (on that machine, at that time).

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