for 8yrs old problem..=)

Code:

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char num[4];
int a,b,c,d;
printf("enter a four digit nuber:");
scanf("%s",num[4]);
a=num[0];
b=num[1];
c=num[2];
d=num[3];
printf("%d",a);
printf("%d",b);
printf("%d",c);
printf("%d",d);

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i input 1234 but it should ouput
a=1
b=2
c=3
d=4
ryt? but it doest output the number that im expecting.. :(

A nudge in the right direction, consider, the character '1' is represented by the character code for '1' which is not equal to 1. Assuming you are using 8 bit characters of a standard type, look up ASCII coding, and the atoi() function.

Quote:

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Originally Posted by rothj0hn

Code:

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char num[4];
scanf("%s",num[4]);

---

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This does not do what you think it does. It should cause a segmentation fault.
You're telling scanf() that you want to put a string at the address stored in num[4], which is beyond the end of the array num[], and is also a char, not a pointer.

I'll leave it to others to explain why making num[] only 4 elements long is a bad idea.

FAQ > How do I... (Level 1) > How do I get a number from the user (C)
FAQ > How do I... (Level 1) > How do I convert an int to a string (char array) (C)

you can try

scanf("%s",&num);

Quote:

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Originally Posted by Belzebuts

you can try

scanf("%s",&num);

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no need of saying & in front of num. b'cose the num is already a pointer to string to the first element

ssharish2005