Understanding Strings

Hy everyone,

I would like to understand a function on strings --> strlen
Below is a code that I took from my teacher where the user inputs a string and prints out the length of the string.

Code:

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


int main()
{
    char str[100];
    int i = 0;
   
    printf("Enter String: ");
    scanf("%s", str);
   
    while (str[i]!='\0')
    {
        i++;
    }
    printf("The String is\n%s\n", str);
    printf("Length of the string: %lu\n", strlen(str));
}

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Now i understand that it returns the count in "int" so my question is:
Let's say i declared

Code:

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int count = 0;

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at the beginning of the code
and then made

Code:

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count = strlen(str);

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why wouldn't i have the same result?
Is there a way to do it also?

Quote:

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Originally Posted by YannB

Let's say i declared

Code:

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int count = 0;

---

at the beginning of the code
and then made

Code:

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count = strlen(str);

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why wouldn't i have the same result?

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You would. It's just a shorthand to pass results of functions as arguments to other functions directly.

Quote:

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Originally Posted by GReaper

You would. It's just a shorthand to pass results of functions as arguments to other functions directly.

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Thank you but when i do that i'm having: implicit conversion loses integer precision: 'unsigned long' to 'int'

could you explain how can i avoid having that?

i fixed it!
but i still don't understand.
from what i understood i need to declare

Code:

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long count = 0;

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why do i need to declare long though?

That is compiler dependent. strlen() actually returns a value of type size_t, which is an implementation-defined unsigned integral type.

If a size_t can represent values that at int cannot, converting it to an int loses information. That is the reason your compiler emits that warning.

It depends on the range of values an int can represent versus the range of values a size_t can represent - both those ranges are implementation defined. Not all compilers will have a problem.

There is also no guarantee that a size_t can be converted to a long without losing information.

The real solution is for count to be given a type of size_t. Don't forget that size_t is an unsigned type.

Line 11, you use no maximum field width with "%s", and so you will overflow str if a word longer than 99 characters is stored. In addition, you have no guarantee from scanf that a string has been stored at all.

Line 18, you tell printf to convert an unsigned long with "%lu", but you give it a size_t.

Below is a fix for these problems. Note that the cast to int is okay as 99, the greatest value that strlen will return, is well within the minimum range that int is guaranteed to have.

Code:

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

int main(void) {
  char s[100];
  if (scanf("%99s", s) == 1)
    printf("%d\n", (int) strlen(s));
  return 0;
}

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If you are wondering what will happen when a word longer than 99 characters is supplied, it will store the first 99 characters and leave the remaining characters on stdin, ready to be read by a following call.

Quote:

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Originally Posted by zyxwvuts

Note that the cast to int is okay as 99, the greatest value that strlen will return, is well within the minimum range that int is guaranteed to have.

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Personally, I think that in principle we should cast from size_t to an unsigned integer type for use with printf, rather than int, even if int will definitely work for the particular case.

C99 has the z length modifier for the size_t type.

Code:

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printf("%zu", strlen(str));

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