how to get the length of the string in one line

i get a string str in a signature
and i dont have an index=0 to use

i cant use strlen because its a function it will give 1 even if the function has 100 cells
i need to do it in one line
len=??1??

Please show what related code you have already, and please please please try to clarify the issue you are having. Your fragmented question provides literally no context to understand your problem.

Code:

---

what(char *str)
{
int len;
len=??
}

---

Ok so why doesn't this work?

Code:

---

len = strlen(str);

---

http://www.cplusplus.com/reference/c...ng/strlen.html

If strlen gives 1 for a char[100], it's because the second character is the null terminator ('\0'). strlen tells the number of characters which precede the null terminator, not how many bytes are allocated to a variable.

For example:

Code:

---

int main(int agrc, char **argv){
    char *foo = "ab\0def";
    printf("The string \"%s\" is %i characters long.", foo, strlen(foo));
}

---

That will print:

Quote:

---

The string "ab" is 2 characters long.

---

because its in external function and we have our array decays to one cell

What?

Sounds like you're trying to implement something to mimic decay mechanism/function of some sort. Might help if you print the block of code that's implementing your logic as I suspect that's where you're going wrong.

Code:

---

what(char *str)
{
  int len;
      len = sizeof(str)/sizeof('a'); // Won't work!
}

---

how to write so it will work
??

i tried

Code:

---

#include <stdio.h>
#include <string.h>
void what(char* str);
int main()
{
        char arr[5]={'a','b','c','d','e'};
    what(arr);
        return 0;
}

void what(char* str)
{
int len;
len=strlen(str);
printf("%d",len);
}

---

it doesnt work

Quote:

---

Originally Posted by transgalactic2

it doesnt work

---

Of course not. You don't have a Null terminator at the end of your string. So strlen will go off searching through memory until it finds one.
if you wrote:

Code:

---

char arr[6]={'a','b','c','d','e', '\0'};

---

That should work.

Quantumpete

Code:

---

#include <stdio.h>
#include <string.h>
void what(char* str);
int main()
{
        char arr[6]={'a','b','c','d','e','\0'};
    what(arr);
        return 0;
}

void what(char* str)
{
int len;
len=strlen(str);
printf("%d",len);
}

---

i changed it it works but i get this warning
c(14) : warning C4267: '=' : conversion from 'size_t' to 'int'
what is size_t why i get this warning

when I run your code, the version with:

Code:

---

char arr[5]={'a','b','c','d','e'};

---

I get the output 7
but with the null terminated one I get the correct output.
What compiler are you using?
I used dev C++.

Quote:

---

Originally Posted by transgalactic2

Code:

---

#include <stdio.h>
#include <string.h>
void what(char* str);
int main()
{
        char arr[6]={'a','b','c','d','e','\0'};
    what(arr);
        return 0;
}

void what(char* str)
{
int len;
len=strlen(str);
printf("%d",len);
}

---

i changed it it works but i get this warning
c(14) : warning C4267: '=' : conversion from 'size_t' to 'int'
what is size_t why i get this warning

---

????

Stop reposting your questions! It's annoying.
As to size_t, it's a datatype. In this case strlen returns size_t (which is probably an unsigned int or something). You declared len as an (signed) int.
to make the program correct, you should either use unsigned int (since strlen cannot return a -ve number) or use size_t len;

QuantumPete

thanks