1. 1 CS 162 Introduction to Computer Science Chapter 5 ASCII to Integer Conversion Herbert G. Mayer, PSU Status 11/9/2014

2. 2 Syllabus #include for getchar() #include for getchar() Convert Decimal Digits to int Convert Decimal Digits to int Account for Sign Account for Sign Check integer Overflow on 16-Bit Check integer Overflow on 16-Bit Maxint on 32-Bit Maxint on 32-Bit

3. 3 Specify a2i() Conversion Reading one input char at a time, assuming the character is a decimal digit ‘0’..’9’ Reading one input char at a time, assuming the character is a decimal digit ‘0’..’9’ Convert this ASCII string of decimal digits to its corresponding integral value Convert this ASCII string of decimal digits to its corresponding integral value For example, the incoming string “123” is converted to the integer hundred-and-twenty-three, i.e. 123 10 For example, the incoming string “123” is converted to the integer hundred-and-twenty-three, i.e. 123 10 Be aware that this input is indeed a sequence of ASCII character, not an integer value Be aware that this input is indeed a sequence of ASCII character, not an integer value So we have to convert the digit ‘5’ for example to the integer value 5, by subtracting from character ‘5’ the value character ‘0’, and the difference is 5 So we have to convert the digit ‘5’ for example to the integer value 5, by subtracting from character ‘5’ the value character ‘0’, and the difference is 5 Int value of ‘5’ is: ‘5’ – ‘0’ Int value of ‘5’ is: ‘5’ – ‘0’

4. 4 Implement a2i() Conversion int a2i( void ) { // a2i int number = 0; int number = 0; while( ( ( c = getchar() ) >= '0' ) && ( c = '0' ) && ( c <= '9' ) ) { number = number * 10 + ( c - '0' ); } //end while } //end while return number; return number; } //end a2i int main() { // main cout << “Enter a decimal number: ” << endl; cout << "The number was:” << a2i() << endl; return 0; } //end main

5. 5 Discuss Code for a2i() Conversion The assumption in the code is that a sequence of decimal digits can be read The assumption in the code is that a sequence of decimal digits can be read However, if not a single digit is entered, the correct result of 0 is generated, due to initialization of number; see “int number = 0;” However, if not a single digit is entered, the correct result of 0 is generated, due to initialization of number; see “int number = 0;” A crucial test is: ( c = getchar() ) >= '0'... A crucial test is: ( c = getchar() ) >= '0'... Here a side-effect happens: A character is read from standard in, and it is tested. Test becomes more complex, but no further input happens Here a side-effect happens: A character is read from standard in, and it is tested. Test becomes more complex, but no further input happens The current number is shifted left by 1 decimal position (multiply by 10) and the digit value is added The current number is shifted left by 1 decimal position (multiply by 10) and the digit value is added

6. 6 Account for Sign in a2i() Before reading one decimal digit at a time, check for optional negative sign ‘-’ or a redundant positive sign ‘+’ Before reading one decimal digit at a time, check for optional negative sign ‘-’ or a redundant positive sign ‘+’ Former ‘-’ will change the resulting value, by inverting the sign Former ‘-’ will change the resulting value, by inverting the sign State of ‘-’ has to be remembered; see variable neg State of ‘-’ has to be remembered; see variable neg The latter ‘+’ is silently skipped The latter ‘+’ is silently skipped Once the sign is handled, proceed as before with the decimal digits that make the number Once the sign is handled, proceed as before with the decimal digits that make the number

7. 7 Implement Sign in a2i() int is_digit( char c ) { // is_digit return ( c >= '0' ) && ( c = '0' ) && ( c <= '9' ); } //end is_digit int a2i0( void ) { // a2i0 int number = 0; int number = 0; int neg = 0; int neg = 0; if ( c == '-' ) { if ( c == '-' ) { neg = 1; neg = 1; c = getchar(); c = getchar(); }else if ( c == '+' ) { }else if ( c == '+' ) { c = getchar(); c = getchar(); } //end if } //end if while ( is_digit( c ) ) { while ( is_digit( c ) ) { number = number * 10 + ( c - '0' ); c = getchar(); } //end while } //end while return neg ? -number : number; return neg ? -number : number; } //end a2i0

8. 8 Discuss Sign in a2i State variable neg is solely used to determine at the end, whether sign inversion happens State variable neg is solely used to determine at the end, whether sign inversion happens Here we us a conditional expression: Here we us a conditional expression: return neg ? -number : number; For the initial state of neg = 0, the decimal computed so far is returned For the initial state of neg = 0, the decimal computed so far is returned But if neg is set to a non-zero state, the sign is inverted, all accomplished in a single conditional expression But if neg is set to a non-zero state, the sign is inverted, all accomplished in a single conditional expression Note separation of the 2 options via the : and ? operators! Note separation of the 2 options via the : and ? operators!

9. 9 Specify Overflow in a2i() on 16-Bit Delicate programming matter, analyzed by sample of a fictitious 16-bit architecture Delicate programming matter, analyzed by sample of a fictitious 16-bit architecture On 16-bit computer the largest possible integer = 32767 On 16-bit computer the largest possible integer = 32767 On a two’s-complement architecture, the largest negative value = -32768 On a two’s-complement architecture, the largest negative value = -32768 Both cases have to be handled, without the overflow actually occurring, so we probe the growing integer value, before the next multiply by 10, MAX10 = 3276 Both cases have to be handled, without the overflow actually occurring, so we probe the growing integer value, before the next multiply by 10, MAX10 = 3276 Also track, whether scanning a negative int literal, in which case the largest next decimal digit for 3276 is 8, versus the positive literal, for which largest next decimal digit fir 3276 is 7 Also track, whether scanning a negative int literal, in which case the largest next decimal digit for 3276 is 8, versus the positive literal, for which largest next decimal digit fir 3276 is 7

10. 10 Specify Overflow in a2i() on 16-Bit Scanning a positive decimal int literal: Scanning a positive decimal int literal: If number scanned so far is less than 3276, which is max 32767 / 10, then scanning 1 more digit is safe, do number = number * 10 + ( c-'0’ ); But if number scanned so far is in critical range 3276, then only some digits are safe, namely ‘0’..’7’ Else there will be positive overflow Scanning a negative decimal int literal: Scanning a negative decimal int literal: If number scanned so far is less than 3276, which is max 32767 / 10, then scanning 1 more digit is safe, do number = number * 10 + ( c-'0’ ); But if number scanned so far is in critical range 3276, then only some digits are safe, namely ‘0’..’8’ Else there will be negative overflow

11. 11 Check for Overflow in a2i() #define MAX 32767 #define MAX10 3276// i.e. integer divide 32767 / 10 = 3276... int a2i1( void ) { // a2i int number = 0; int number = 0; int neg = 0; int neg = 0; if ( c == '-' ) { if ( c == '-' ) { neg = 1; neg = 1; c = getchar(); c = getchar(); }else if ( c == '+' ) { }else if ( c == '+' ) { c = getchar();// keep neg = 0 c = getchar();// keep neg = 0 } //end if } //end if while ( is_digit( c ) ) { while ( is_digit( c ) ) { if ( number > MAX10 ) { printf( "Overflow 1. max = %d\n", MAX ); printf( "Overflow 1. max = %d\n", MAX ); }else if ( number == MAX10 ) { if ( c == '9' ) { if ( c == '9' ) { printf( "Overflow 2. max = %d\n", MAX ); }else if ( c == '8' ) { }else if ( c == '8' ) { if ( neg ) { number = -32768; number = -32768;}else{ printf( "Overflow 3. max = %d\n", MAX ); printf( "Overflow 3. max = %d\n", MAX ); } //end if }else{ }else{ number = number * 10 + ( c - '0' ); } //end if } //end if}else{ number = number * 10 + ( c - '0' ); number = number * 10 + ( c - '0' ); } //end if } //end if c = getchar(); } //end while } //end while return neg ? -number : number; return neg ? -number : number; } //end a2i

12. 12 Discuss Overflow in a2i The critical number = number * 10 + next digit is performed only in safe state: The critical number = number * 10 + next digit is performed only in safe state: For positive state, if the next digit after 3276 is ‘0’..’7’ one more iteration of the multiply is safe; yielding <= 32767 For positive state, if the next digit after 3276 is ‘0’..’7’ one more iteration of the multiply is safe; yielding <= 32767 For negative state, if the next digit after 3276 is ‘0’..’8’ one more iteration of the multiply is safe; yielding >= -32768 For negative state, if the next digit after 3276 is ‘0’..’8’ one more iteration of the multiply is safe; yielding >= -32768 Similarly on a 32-bit or today’s 64-bit computer; the scanner for integer literals needs to be safe, and detect integer overflow before it happens Similarly on a 32-bit or today’s 64-bit computer; the scanner for integer literals needs to be safe, and detect integer overflow before it happens Common to have computers ignore integer overflow, so it may be safe to simply experience it, but the numeric value has to be safely scanned Common to have computers ignore integer overflow, so it may be safe to simply experience it, but the numeric value has to be safely scanned

13. 13 Maxint on 32-Bit #include #include main() { // main int expo;// shift 1 32 times int expo;// shift 1 32 times unsigned long int pow = 1;// multiply repeatedly by 2 unsigned long int pow = 1;// multiply repeatedly by 2 for ( expo = 1; expo <= 32; expo++ ) { for ( expo = 1; expo <= 32; expo++ ) { pow = pow * 2; pow = pow * 2; cout << " 2**" << expo << " = " << pow-1 << endl; cout << " 2**" << expo << " = " << pow-1 << endl; } // end for } // end for cout << 0+4294967295 << endl; cout << 0+4294967295 << endl; cout << 0-4294967295 << endl; cout << 0-4294967295 << endl; } //end main

14. 14 Maxint Output 2**1 = 1 2**1 = 1 2**2 = 3 2**2 = 3 2**3 = 7 2**3 = 7...... 2**14 = 16383 2**14 = 16383 2**15 = 32767 2**15 = 32767 2**16 = 65535 2**16 = 65535 2**17 = 131071 2**17 = 131071 2**18 = 262143 2**18 = 262143 2**19 = 524287 2**19 = 524287 2**20 = 1048575 2**20 = 1048575 2**21 = 2097151 2**21 = 2097151 2**22 = 4194303 2**22 = 4194303 2**23 = 8388607 2**23 = 8388607 2**24 = 16777215 2**24 = 16777215 2**25 = 33554431 2**25 = 33554431 2**26 = 67108863 2**26 = 67108863 2**27 = 134217727 2**27 = 134217727 2**28 = 268435455 2**28 = 268435455 2**29 = 536870911 2**29 = 536870911 2**30 = 1073741823 2**30 = 1073741823 2**31 = 2147483647 2**31 = 2147483647 2**32 = 4294967295 2**32 = 429496729542949672951