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From: David Brown <david.brown@hesbynett.no>
Newsgroups: comp.lang.c
Subject: Re: "undefined behavior"?
Date: Thu, 13 Jun 2024 15:28:20 +0200
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On 13/06/2024 02:19, Janis Papanagnou wrote:
> On 13.06.2024 00:22, Keith Thompson wrote:
>>
>> This:
>>      char outliers[100] = "";
>> initializes all 100 elements to zero.  So does this:
>>      char outliers[100] = { '\0' };
>> Any elements or members not specified in an initializer are set to zero.
> 
> Oops! This surprised me. (But you are right.) The overhead isn't
> [syntactically] obvious, but I'm anyway always setting a single
> '\0' character if I want to store strings in a 'char[]' and have
> it initialized to an empty string (like below).
> 
>> If you want to set an array's 0th element to 0 and not waste time
>> initializing the rest, you can assign it separately:
>>      char outliers[100];
>>      outliers[0] = '\0';
>> or
>>      char outliers[100];
>>      strcpy(outliers, "");
>> though the overhead of the function call is likely to outweigh the
>> cost of initializing the array.
> 
> It wouldn't occur to me to use the strcpy() function, but is the
> function call really that expensive in C ?
> 

That depends on your toolchain.

If you are using a Windows-based compiler with an external DLL for the C 
library and the compiler doesn't handle the strcpy() directly, then it 
can be quite a lot of overhead.  You have the call to the DLL, which 
involves a few steps of indirection.  The library strcpy() may be 
optimised for handling large strings, and may save and restore a lot of 
registers (such as SIMD vector registers).

If you are using a compiler (whatever the platform) that optimises 
"strcpy", it will generate identical code to "outliers[0] = '\0';".