Convert a numeric string to an integer

Question: compile a C function that converts a given string to an integer.

[This program is compiled in Dev C ++ 4.9.9.2]

The following program only considers the decimal string.

Int strtoint (char * Str)

{

Int value = 0;

Int Sign = 1;

If (* STR = '-')

{

Sign =-1;

STR ++;

}

While (* Str)

{

Value = value * 10 + * str-'0 ';

STR ++;

}

Return sign * value;

}

Int main ()

{

Printf ("number = % d/N", strtoint ("1234567 "));

Printf ("number = % d/N", strtos ("-1234567 "));

Printf ("number = % d/N", strtoint ("2147483647 "));

Printf ("number = % d/N", strtos ("-2147483647 "));

System ("pause ");

Return 0;

}

The following program considers octal, decimal, and hexadecimal strings.

Int strtoint (char * Str)

{

Int value = 0;

Int Sign = 1;

Int Radix;

 

If (* STR = '-')

{

Sign =-1;

STR ++;

}

If (* STR = '0' & (* (STR + 1) = 'X' | * (STR + 1) = 'X '))

{

Radix = 16;

STR + = 2;

}

Else if (* STR = '0 ')

{

Radix = 8;

STR ++;

}

Else

Radix = 10;

While (* Str)

{

If (Radix = 16)

{

If (* STR> = '0' & * STR <= '9 ')

Value = value * Radix + * str-'0 ';

Else

Value = value * Radix + (* STR | 0x20)-'A' + 10;

}

Else

Value = value * Radix + * str-'0 ';

STR ++;

}

Return sign * value;

}

Int main ()

{

Printf ("decimal string translation! /N ");

Printf ("/" 1234567/"= % d/N", strtoint ("1234567 "));

Printf ("/"-1234567/"= % d/N", strtoint ("-1234567 "));

Printf ("/" 2147483647/"= % d/N", strtoint ("2147483647 "));

Printf ("/"-2147483647/"= % d/N", strtoint ("-2147483647 "));

Printf ("/nhex string translation! /N ");

Printf ("/" 0x200/"= % d/N", strtoint ("0x200 "));

Printf ("/"-0x200/"= % d/N", strtoint ("-0x200 "));

Printf ("/" 0x7fffffff/"= % d/N", strtoint ("0x7fffffff "));

Printf ("/"-0x7fffffff/"= % d/N", strtoint ("-0x7fffffff "));

Printf ("/noctal string translation! /N ");

Printf ("/" 0123/"= % d/N", strtoint ("0123 "));

Printf ("/"-0123/"= % d/N", strtoint ("-0123 "));

System ("pause ");

Return 0;

}