String conversion to int - c

Question

I have a pointer lpBegin pointing to a string "1234". Now i want this string compare to an uint how can i make this string to unsigned integer without using scanf? I know the string number is 4 characters long.

Answer 1

You will have to use the atoi function. This takes a pointer to a char and returns an int.
const char *str = "1234";
int res = atoi(str); //do something with res
As said by others and something I didn't know, is that atoi is not recommended because it is undefined what happens when a formatting error occurs. So better use strtoul as others have suggested.

Answer 2

Definitely atoi() which is easy to use.
Don't forget to include stdlib.h.

Answer 3

You can use the strtoul() function. strtoul stands for "String to unsigned long":
#include <stdio.h>
#include <stdlib.h>
int main()
{
char lpBegin[] = "1234";
unsigned long val = strtoul(lpBegin, NULL, 10);
printf("The integer is %ul.", val);
return 0;
}
You can find more information here: http://www.cplusplus.com/reference/clibrary/cstdlib/strtoul/

Answer 4

You could use strtoul(lpBegin), but this only works with zero-terminated strings.
If you don't want to use stdlib for whatever reason and you're absolutely sure about the target system(s), you could do the number conversion manually.
This one should work on most systems as long as they are using single byte encoding (e.g. Latin, ISO-8859-1, EBCDIC). To make it work with UTF-16, just replace the 'char' with 'wchar_t' (or whatever you need).
unsigned long sum = (lpbegin[0] - '0') * 1000 +
(lpbegin[1] - '0') * 100 +
(lpbegin[2] - '0') * 10 +
(lpbegin[3] - '0');
or for numbers with unknown length:
char* c = lpBegin;
unsigned long sum = 0;
while (*c >= '0' && *c <= '9') {
sum *= 10;
sum += *c - '0';
++c;
}

Answer 5

I think you look for atoi()
http://www.elook.org/programming/c/atoi.html

Answer 6

strtol is better than atoi with better error handling.

Answer 7

You should use the strtoul function, "string to unsigned long". It is found in stdlib.h and has the following prototype:
unsigned long int strtoul (const char * restrict nptr,
char ** restrict endptr,
int base);
nptr is the character string.
endptr is an optional parameter giving the location of where the function stopped reading valid numbers. If you aren't interested of this, pass NULL in this parameter.
base is the number format you expect the string to be in. In other words, 10 for decimal, 16 for hex, 2 for binary and so on.
Example:
#include <stdlib.h>
#include <stdio.h>
int main()
{
const char str[] = "1234random rubbish";
unsigned int i;
const char* endptr;
i = strtoul(str,
(char**)&endptr,
10);
printf("Integer: %u\n", i);
printf("Function stopped when it found: %s\n", endptr);
getchar();
return 0;
}
Regarding atoi().
atoi() internally just calls strtoul with base 10. atoi() is however not recommended, since the C standard does not define what happens when atoi() encounters a format error: atoi() can then possibly crash. It is therefore better practice to always use strtoul() (and the other similar strto... functions).

Answer 8

If you're really certain the string is 4 digits long, and don't want to use any library function (for whatever reason), you can hardcode it:
const char *lpBegin = "1234";
const unsigned int lpInt = 1000 * (lpBegin[0] - '0') +
100 * (lpBegin[1] - '0') +
10 * (lpBegin[2] - '0') +
1 (lpBegin[3] - '0');
Of course, using e.g. strtoul() is vastly superior so if you have the library available, use it.