Wrong result from the conversion in an array

Question

Let's say that I have

char number[2] = "2";

In my code I get number 2 as a string that's why i have char. Now with the usage of atoi I convert this char to int

int conv_number;
conv_number = atoi(number);
printf("Result : %d\n", conv_number);

which returns me Result : 2. Now I want to put this value in an array and print the result of the array.So I wrote

int array[] = {conv_number};
printf("%d\n",array);

Unfortunately my result is not 2 but -1096772864. What am I missing;

array is not actually an address of the array, rather decays to a pointer to the first element in the array which is equal to the address of the array. Pls correct if am wrong — Vivek Maran
– Vivek Maran, Commented Jul 17, 2013 at 11:34
This: `int array[] = {conv_number};' also makes a one-element array, which isn't very useful. — Crowman
– Crowman, Commented Jul 17, 2013 at 15:11

Zeta · Accepted Answer · 2013-07-17 14:38:25Z

9

You're missing that your array is int[] not int, which is the expected second argument for printf when you use the digit format parameter %d.

Use printf("%d\n",array[0]) instead, since you want to access the first value in your array.

Further explanation

In this circumstances array in your printf expression behaves as int*. You would get the same result if you were to use printf("%d\n",&array[0]), aka the address of the first element. Note that if you're really interested in the address use the %p format specifier instead.

edited Jul 17, 2013 at 14:38

answered Jul 17, 2013 at 11:10

Zeta

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Comments

Will · Accepted Answer · 2013-07-17 11:34:40Z

2

in the expression printf("%d\n",array);, array is an array (obviously) of int, which is similar to an int*. the value of array is not the value of the first cell (eg. array[0]) but decays to the array's address.

If you run your code multiple times, you'll probably have different values (the array location will vary from one run to an other)

You have to write : printf("%d\n",array[0]); which is equivalent to printf("%d\n",*array);

edited Jul 17, 2013 at 11:34

Will

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answered Jul 17, 2013 at 11:15

johan d

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Comments

Chinna · Accepted Answer · 2013-07-17 11:14:50Z

1

You are printing base address of your array. Instead of that you need to print value at that array address. like,

printf("%d",array[0]);

answered Jul 17, 2013 at 11:14

Chinna

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1 Comment

Maroun Over a year ago

Actually printing the address should done with %p and not %d.

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