Array copying C programming

Question

I have the code as:

mid = 3;
for(i=0;i<9;i++)
{
   if(i == 9-mid)
      num[i] = mid;
   else
      num[i] = 0;
   printf("%d", num[i]);
}

which gives the result of "000000300".

What I try to do is to store "000000300" as an element of another array, i.e.

unsigned int array[0] = 000000300;

Any ideas of how to do this in C? Thanks~

What is `num` declared as? Also, do you want to store the integer `300` or the string `"000000300"`? — DanielGibbs, Nov 18 '14 at 23:14
You would have to store this with characters otherwise it's 300 as a int — Rizier123, Nov 18 '14 at 23:14
@Rizier123 how to store this? Could you please explain more? — Shumin Xu, Nov 18 '14 at 23:18
You are printing **each element in `num`, not just a single number** (you have a `for` loop wrapping the `printf`). You can either `memcpy` the entire array, or use another `for` loop for `element-by-element` copy (assignment). — David C. Rankin, Nov 18 '14 at 23:23
Please note that 00000300 is an octal constant whose decimal equivalent is 192. — Bob Jarvis - Слава Україні, Nov 18 '14 at 23:39

score -1 · Answer 1 · answered Nov 18 '14 at 23:24

-1

If you want to copy the calculated string "000000300" you will need to allocate some memory and store it in a char * array:

// num is a char array containing "000000300".
char *stored = (char *)malloc(strlen(num) + 1);
if (stored == NULL) {
    // This means that there is no memory available.
    // Unlikely to happen on modern machines.
}
strcpy(stored, num);

answered Nov 18 '14 at 23:24

DanielGibbs

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Any way to scan the num array and pass the value "000000300" to an unsigned int array? – Shumin Xu Nov 18 '14 at 23:45
You could try and parse the string as an int (see http://stackoverflow.com/a/3421555/343486) but having leading 0's will cause problems as it will be interpreted as an octal number. – DanielGibbs Nov 19 '14 at 13:40

Array copying C programming

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