When using atoi in C I am trying to convert a char array of numbers to an int. I have leading 0's on my number though and they are not preserved when I print the number out later.
char num[] = "00905607";
int number;
number = atoi(num);
printf("%d", number);
The output from this will be 905607 and I would like it to be 00905607.
Any ideas?
Use strtol instead. It allows you to specify the base.
You can do padding on your printf() so if you wanted every output to be 8 characters long you would use
printf("%08d", number);
That code is working properly, if it works at all. The integer is 905607... leading zeros don't exist in a mathematical sense.
You have a lot of issues with that code. You're declaring your string improperly, and you're not printing the converted number. Also, if you were doing printf(number); you'd need to use a format string. If you'd like to have leading spaces in that, you can use a width specifier.
Don't use atoi. Use strtol with a base of 10.
In general, there's no reason to use atoi in modern code.
Yes, when you want to display ints you have to format them as strings again. An integer is just a number, it doesn't contain any information on how to display it. Luckily, the printf-function already contains this, so that would be something like
printf( "%08d", num);
Most answers assume you want 8 digits while this is not exactly what you requested.
If you want to keep the number of leading zeros, you're probably better keeping it as a string, and convert it with strtol for calculations.
You could use sscanf, and provide '%d' as your format string.
If you don't want that behavior; don't use atoi.
Perhaps sscanf with a %d format?
You could also count the numbers of numbers in the string and create a new one with leading zeroes.
Or check out all the wonderful format tags for printf and use one to pad with zeroes. Here for example are lot of them: http://www.cplusplus.com/reference/clibrary/cstdio/printf/
