Will strlen be calculated multiple times if used in a loop condition?

Question

I'm not sure if the following code can cause redundant calculations, or is it compiler-specific?

for (int i = 0; i < strlen(ss); ++i)
{
    // blabla
}

Will strlen() be calculated every time when i increases?

Answer 1

Yes, strlen() will be evaluated on each iteration. It's possible that, under ideal circumstances, the optimiser might be able to deduce that the value won't change, but I personally wouldn't rely on that.

I'd do something like

for (int i = 0, n = strlen(ss); i < n; ++i)

or possibly

for (int i = 0; ss[i]; ++i)

as long as the string isn't going to change length during the iteration. If it might, then you'll need to either call strlen() each time, or handle it through more complicated logic.

Answer 2

Yes, every time you use the loop. Then it will every time calculate the length of the string. so use it like this:

char str[30];
for ( int i = 0; str[i] != '\0'; i++)
{
//Something;
}

In the above code str[i] only verifies one particular character in the string at location i each time the loop starts a cycle, thus it will take less memory and is more efficient.

See this Link for more information.

In the code below every time the loop runs strlen will count the length of the whole string which is less efficient, takes more time and takes more memory.

char str[];
for ( int i = 0; i < strlen(str); i++)
{
//Something;
}

Answer 3

A good compiler may not calculate it every time, but I don't think you can be sure, that every compiler does it.

In addition to that, the compiler has to know, that strlen(ss) does not change. This is only true if ss is not changed in for loop.

For example, if you use a read-only function on ss in for loop but don't declare the ss-parameter as const, the compiler cannot even know that ss is not changed in the loop and has to calculate strlen(ss) in every iteration.

Answer 4

If ss is of type const char * and you're not casting away the constness within the loop the compiler might only call strlen once, if optimizations are turned on. But this is certainly not behavior that can be counted upon.

You should save the strlen result in a variable and use this variable in the loop. If you don't want to create an additional variable, depending on what you're doing, you may be ale to get away with reversing the loop to iterate backwards.

for( auto i = strlen(s); i > 0; --i ) {
  // do whatever
  // remember value of s[strlen(s)] is the terminating NULL character
}

Answer 5

Formally yes, strlen() is expected to be called for every iteration.

Anyway I do not want to negate the possibility of the existance of some clever compiler optimisation, that will optimise away any successive call to strlen() after the first one.

Answer 6

The predicate code in it's entirety will be executed on every iteration of the for loop. In order to memoize the result of the strlen(ss) call the compiler would need to know that at least

1. The function strlen was side effect free
2. The memory pointed to by ss doesn't change for the duration of the loop

The compiler doesn't know either of these things and hence can't safely memoize the result of the first call

Answer 7

Yes. strlen will be calculated everytime when i increases.

If you didn't change ss with in the loop means it won't affect logic otherwise it will affect.

It is safer to use following code.

int length = strlen(ss);

for ( int i = 0; i < length ; ++ i )
{
 // blabla
}

Answer 8

Yes, strlen(ss) will be calculated every time the code runs.

Answer 9

Yes, the strlen(ss) will calculate the length at each iteration. If you are increasing the ss by some way and also increasing the i; there would be infinite loop.

Answer 10

Yes, the strlen() function is called every time the loop is evaluated.

If you want to improve the efficiency then always remember to save everything in local variables... It will take time but it's very useful ..

You can use code like below:

String str="ss";
int l = strlen(str);

for ( int i = 0; i < l ; i++ )
{
    // blablabla
}

Answer 11

Not common nowadays but 20 years ago on 16 bit platforms, I'd recommend this:

for ( char* p = str; *p; p++ ) { /* ... */ }

Even if your compiler isn't very smart in optimization, the above code can result in good assembly code yet.

Answer 12

Yes. The test doesn't know that ss doesn't get changed inside the loop. If you know that it won't change then I would write:

int stringLength = strlen (ss); 
for ( int i = 0; i < stringLength; ++ i ) 
{
  // blabla 
} 

Answer 13

Arrgh, it will, even under ideal circumstances, dammit!

As of today (January 2018), and gcc 7.3 and clang 5.0, if you compile:

#include <string.h>

void bar(char c);

void foo(const char* __restrict__ ss) 
{
    for (int i = 0; i < strlen(ss); ++i) 
    {
        bar(*ss);
    }
}    

So, we have:

• ss is a constant pointer.
• ss is marked __restrict__
• The loop body cannot in any way touch the memory pointed to by ss (well, unless it violates the __restrict__).

and still, both compilers execute strlen() every single iteration of that loop. Amazing.

This also means the allusions/wishful thinking of @Praetorian and @JaredPar doesn't pan out.

Answer 14

YES, in simple words. And there is small no in rare condition in which compiler is wishing to, as an optimization step if it finds that there is no changes made in ss at all. But in safe condition you should think it as YES. There are some situation like in multithreaded and event driven program, it may get buggy if you consider it a NO. Play safe as it is not going to improve the program complexity too much.

Answer 15

Yes.

strlen() calculated everytime when i increases and does not optimized.

Below code shows why the compiler should not optimize strlen().

for ( int i = 0; i < strlen(ss); ++i )
{
   // Change ss string.
   ss[i] = 'a'; // Compiler should not optimize strlen().
}

Answer 16

We can easily test it :

char nums[] = "0123456789";
size_t end;
int i;
for( i=0, end=strlen(nums); i<strlen(nums); i++ ) {
    putchar( nums[i] );
    num[--end] = 0;
}

Loop condition evaluates after each repetition, before restarting the loop .

Also be careful about the type you use to handle length of strings . it should be size_t which has been defined as unsigned int in stdio. comparing and casting it to int might cause some serious vulnerability issue.

Answer 17

well, I noticed that someone is saying that it is optimized by default by any "clever" modern compiler. By the way look at results without optimization. I tried:
Minimal C code:

#include <stdio.h>
#include <string.h>

int main()
{
 char *s="aaaa";

 for (int i=0; i<strlen(s);i++)
  printf ("a");
 return 0;
}

My compiler: g++ (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3
Command for generation of assembly code: g++ -S -masm=intel test.cpp

Gotten assembly code at the output:
    ...
    L3:
mov DWORD PTR [esp], 97
call    putchar
add DWORD PTR [esp+40], 1
    .L2:
     THIS LOOP IS HERE
    **<b>mov    ebx, DWORD PTR [esp+40]
mov eax, DWORD PTR [esp+44]
mov DWORD PTR [esp+28], -1
mov edx, eax
mov eax, 0
mov ecx, DWORD PTR [esp+28]
mov edi, edx
repnz scasb</b>**
     AS YOU CAN SEE it's done every time
mov eax, ecx
not eax
sub eax, 1
cmp ebx, eax
setb    al
test    al, al
jne .L3
mov eax, 0
     .....

Answer 18

Elaborating on Prætorian's answer I recommend the following:

for( auto i = strlen(s)-1; i > 0; --i ) {foo(s[i-1];}

• auto because you don't want to care about which type strlen returns. A C++11 compiler (e.g. gcc -std=c++0x, not completely C++11 but auto types work) will do that for you.
• i = strlen(s) becuase you want to compare to 0 (see below)
• i > 0 because comparison to 0 is (slightly) faster that comparison to any other number.

disadvantage is that you have to use i-1 in order to access the string characters.