How can i get the size of my char * variable without using a for loop?

Question

char *foo(char *dest, const char *src) {
    size_t i;                      
    for (i = 0; dest[i] != '\0'; i++);

Here, I'm iterating to get the size of dest. In this case I’d input "hello " into dest, which is a size of 6. When I try to use sizeof(dest) I get 4 as the returned value. I want to be able to get the size of the content that's inside of dest without using this for loop.

char *foo(char *dest, const char *src) {
    while (*dest != '\0') dest++;           /* increment the length of dest's pointer*/

EDIT:: i'd like to take a moment to show that i was able to get around finding the length directly.

This is all part of a strcat program. The requirement was to not use [ ] brackets to access or move around in memory.

char *strcat(char *dest, const char *src) {
    while (*dest != '\0') dest++;           /* increment the length of dest's pointer*/
    while (*src != '\0')                    /* we will be incrementing up through src*/
        *dest++ = *src++;                   /* while this is happening we are appending
                                             * letter by letter onto the variable dest
                                             */
    *(dest++) = ' ';                        /* increment up one in memory and add a space */
    *(dest++) = '\0';                       /* increment up one in memory and add a null
                                             * termination at the end of our variable dest
                                             */
    return dest;                            /* return the final output */
}

Also see http://stackoverflow.com/questions/492384/how-to-find-the-sizeofa-pointer-pointing-to-an-array — Lundin, Feb 05 '16 at 07:33

score 6 · Accepted Answer · answered Feb 05 '16 at 05:26

6

With null terminated strings you have to iterate over each character to work out the length. Even if you use the strlen() it will be doing what your loop does.

answered Feb 05 '16 at 05:26

Matthew Panetta

88
4

score 0 · Answer 2 · answered Feb 05 '16 at 05:24

0

You're looking for strlen(). But note that it's likely implemented with that same loop.

answered Feb 05 '16 at 05:24

Bill Lynch

80,138
16
128
173

score 0 · Answer 3 · answered Feb 05 '16 at 05:31

Since the type of dest in your function is char const*, sizeof(dest) is the same as sizeof(char const*), i.e. the size of a pointer. The fact that you get 4 when you use sizeof(dest) indicates that sizeof of a pointer in your platform is 4.

The only way to get the length of the string is to count the characters until you encounter the null character. This is most likely what strlen does too.

score 0 · Answer 4 · answered Feb 05 '16 at 05:31

In C, strings are stored as character arrays terminated in a \0.

If you want to get the number of characters in an array, you have to loop over this array, you cannot get around it.

However, you can store the size in a structure,

typedef struct
{ 
  int size;
  char *data;
}String;

Then, you have to make wrapper functions to write to this String, read from this String, and update the data.

This can be useful if you have lots of reads of size and not many updates or writes (or zero writes for constants).

But generally, the for loop is the better solution.

score -1 · Answer 5 · answered Feb 05 '16 at 05:48

-1

yeah!! u can get size without using for loop. introduce library and strlen(dest)+1 will be size of your array.cout<< dest; will surely give your array but sizeof(dest) will not give size of array. I am also confused why that happens.

answered Feb 05 '16 at 05:48

Amrinder Singh

1
1

`sizeof(dest)` won't give you the size of an array because it gives you the size of `dest` (duh) and `dest` is not an array. – user253751 Feb 05 '16 at 05:53
but why cout< – Amrinder Singh Feb 05 '16 at 05:57
That question doesn't even make sense. – user253751 Feb 05 '16 at 06:28

How can i get the size of my char * variable without using a for loop?

5 Answers5