How to covert char[i] to int?

I have a number (12345678)
stored in
char[0]=1
char[1]=2
char[2]=3
char[3]=4
and so on...........I would like to convert this and stored it as one interger value. That is
int t=12345678.
How can I do this?

Or, you can do like this too,char[] ch = {'1','2','3','4','5','6','7','8'};
     int in = 0;
     int len = ch.length;
     for(int i=0;i<len;i++) {
          in += Character.getNumericValue(ch) * Math.pow(10,(double)(len-i-1));
     System.out.println("Value is : "+in);

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Is this an error in VC++, or am I missing something? The following code should give two warnings for assigning to a smaller type, but only gives one: (compiled as 32 bit)
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http://en.wikipedia.org/wiki/Courier_(typeface)

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johnsold wrote:
PKIM,
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<begging removed by moderator>
Gaur Mayank
Edited by: Gaur Mayank on Nov 5, 2011 10:40 AM
Edited by: Thomas Zloch on Nov 6, 2011 3:59 PM

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