6610 बाइट्स (अघोषित)
"अच्छा लड़का" सी कार्यक्रम जो सभी चुनौती मानदंडों को पूरा करता है। नकारात्मक संख्याओं के लिए 10 एस पूरक का उपयोग करता है। भी शामिल है, एक परीक्षण दोहन और परीक्षण के मामले।
/*
Read input from STDIN. The input must conform to VALIDINPUT:
NONZERODIGIT = "1" / "2" / "3" / "4" / "5" / "6" / "7" / "8" / "9"
POSITIVENUMBER = NONZERODIGIT *98DIGIT
NEGATIVENUMBER = "-" POSITIVENUMBER
NUMBER = NEGATIVENUMBER / POSITIVENUMBER / "0"
VALIDINPUT = NUMBER SP NUMBER *1LF EOF
Check syntax of input. If input is correct, add the two numbers and print
to STDOUT.
LSB => least significant byte
MSB => most significant byte
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <stdbool.h>
#include <assert.h>
#define NUL ('\0')
/*
maximum characters in VALIDINPUT:
'-' 1
POSITIVENUMBER MAXDIGITS
' ' 1
'-' 1
POSITIVENUMBER MAXDIGITS
LF 1
*/
#define MAXDIGITS (99)
#define MAXVALIDINPUT (2*MAXDIGITS+4)
void die() { printf("E\n"); exit(1); }
/*
Add two NUMBERs and print the result to STDOUT. The NUMBERS have
been separated into POSITIVENUMBERS and sign information.
Arguments:
first - pointer to LSB of 1st POSITIVENUMBER
firstSize - size of 1st POSITIVENUMBER
firstNegative - is 1st # negative?
second - pointer to LSB of 2nd POSITIVENUMBER
secondSize - size of 2nd POSITIVENUMBER
secondNegative - is 2nd # negative?
carry - carry from previous place?
Returns:
sum[] - sum
addNUMBERs() - carry to next place?
- Don't use complementDigit(popDigit(p,s),n). Side-effects generate two pops.
*/
#define popDigit(p,s) ((s)--,(*(p++)-'0'))
#define complementDigit(c,n) ((n) ? 9-(c) : (c))
#define pushSum(c) (*(--sumPointer)=(c))
#define openSum() (pushSum(NUL))
#define closeSum() ;
char sum[MAXVALIDINPUT];
char *sumPointer = sum+sizeof(sum);
int addNUMBERs(char *first, int firstSize, bool firstNegative,
char *second, int secondSize, bool secondNegative,
int previousCarry) {
int firstDigit, secondDigit;
int mySum;
int myCarry;
/*
1st half of the problem.
Build a stack of digits for "first" and "second"
numbers. Each recursion of addNUMBERs() contains one digit
of each number for that place. I.e., the 1st call holds
the MSBs, the last call holds the LSBs.
If negative, convert to 10s complement.
*/
assert((firstSize > 0) && (secondSize > 0));
if (firstSize > secondSize) {
firstDigit = popDigit(first, firstSize);
firstDigit = complementDigit(firstDigit, firstNegative);
secondDigit = 0;
} else if (secondSize > firstSize) {
firstDigit = 0;
secondDigit = popDigit(second, secondSize);
secondDigit = complementDigit(secondDigit, secondNegative);
} else {
// same size
firstDigit = popDigit(first, firstSize);
firstDigit = complementDigit(firstDigit, firstNegative);
secondDigit = popDigit(second, secondSize);
secondDigit = complementDigit(secondDigit, secondNegative);
}
// recursion ends at LSB
if ((firstSize == 0) && (secondSize == 0)) {
// if negative, add 1 to complemented LSB
if (firstNegative) {
firstDigit++;
}
if (secondNegative) {
secondDigit++;
}
myCarry = previousCarry;
} else {
myCarry = addNUMBERs(first, firstSize, firstNegative,
second, secondSize, secondNegative,
previousCarry);
}
/*
2nd half of the problem.
Sum the digits and save them in first[].
*/
mySum = firstDigit + secondDigit + ((myCarry) ? 1 : 0);
if ((myCarry = (mySum > 9))) {
mySum -= 10;
}
pushSum(mySum + '0');
return(myCarry);
}
// Handle the printing logic.
void addAndPrint(char *first, int firstSize, bool firstNegative,
char *second, int secondSize, bool secondNegative,
int previousCarry) {
openSum();
addNUMBERs(first, firstSize, firstNegative,
second, secondSize, secondNegative,
previousCarry)
closeSum();
if (*sumPointer<'5') {
// it's positive
for (; *sumPointer=='0'; sumPointer++) {} // discard leading 0s
// if all zeros (sumPointer @ NUL), back up one
sumPointer -= (*sumPointer == NUL) ? 1 : 0;
printf("%s\n", sumPointer);
} else {
// it's negative
char *p;
// discard leading 0s (9s in 10s complement)
for (; *sumPointer=='9' && *sumPointer; sumPointer++) {}
// if -1 (sumPointer @ EOS), back up one
sumPointer -= (*sumPointer == NUL) ? 1 : 0;
for (p=sumPointer; *p; p++) {
*p = '0' + ('9' - *p); // uncomplement
// special handling, +1 for last digit
*p += (*(p+1)) ? 0 : 1;
}
printf("-%s\n", sumPointer);
}
return;
}
/*
Lex a number from STDIN.
Arguments:
bufferPointer - pointer to a pointer to a buffer, use as
**buffer = c; // put "c" in the buffer
*buffer += 1; // increment the buffer pointer
(*buffer)++; // also increments the buffer pointer
All sorts of side-effects:
- getc(stdin)
- ungetc(...,stdin)
- modifies value of **bufferPointer
- modifies value of *bufferPointer
Returns:
lexNUMBER() - number of bytes added to *bufferPointer,
*1 if POSITIVENUMBER,
*-1 if NEGATIVENUMBER
*bufferPointer - points to the LSB of the number parsed + 1
*/
#define pushc(c) (*((*bufferPointer)++)=c)
bool lexNUMBER(char **bufferPointer) {
char c;
int size = 0;
bool sign = false;
/* lex a NUMBER */
if ((c=getchar()) == '0') {
pushc(c);
c = getchar();
size++;
} else {
if (c == '-') {
sign = true;
c = getchar();
// "-" isn't a digit, don't add to size
}
if (c == '0') {
die();
}
for (size=0; isdigit(c); size++) {
if (size >= MAXDIGITS) {
die();
}
pushc(c);
c = getchar();
}
}
if (size < 1) {
die();
}
ungetc(c,stdin); // give back unmatched character
return (sign);
}
int main() {
int c;
char buffer[MAXVALIDINPUT];
char *bufferPointer;
char *first, *second;
int firstSize, secondSize;
bool firstNegative, secondNegative;
bufferPointer = buffer + 1; // hack, space for leading digit
// parse 1st number
first = bufferPointer;
firstNegative = lexNUMBER(&bufferPointer);
firstSize = bufferPointer - first;
*(bufferPointer++) = NUL; // hack, space for EOS
bufferPointer++; // hack, space for leading digit
// parse separating blank
if ((c=getchar()) != ' ') {
die();
}
// parse 2nd number
second = bufferPointer;
secondNegative = lexNUMBER(&bufferPointer);
secondSize = bufferPointer - second;
*(bufferPointer++) = NUL; // hack, space for EOS
// parse end of input
c = getchar();
if (! ((c == EOF) || ((c == '\n') && ((c=getchar()) == EOF))) ) {
die();
}
// Some very implementation-specific massaging.
*(--first) = '0'; // prefix with leading 0
*(first+(++firstSize)) = NUL; // add EOS
*(--second) = '0'; // prefix with leading 0
*(second+(++secondSize)) = NUL; // add EOS
// add and print two arbitrary precision numbers
addAndPrint(first, firstSize, firstNegative,
second, secondSize, secondNegative, false);
return(0);
}
यहां थोड़ा परीक्षण हार्नेस और कुछ परीक्षण मामलों को आरंभ करने के लिए किया गया है। बेझिझक पर्ल के अत्यधिक उपयोग को फाड़ दें। जिस सिस्टम पर इसे विकसित किया गया था, उसमें आधुनिक बैश नहीं था।
#!/bin/bash
#
# testharness.sh
#
# Use as: bash testharness.sh program_to_be_tested < test_data
#
# Each line in the test data file should be formatted as:
#
# INPUT = bash printf string, must not contain '"'
# OUTPUT = perl string, must not contain '"'
# (inserted into the regex below, use wisely)
# TESTNAME = string, must not contain '"'
# GARBAGE = comments or whatever you like
# INPUTQUOTED = DQUOTE INPUT DQUOTE
# OUTPUTQUOTED = DQUOTE OUTPUT DQUOTE
# TESTQUOTED = DQUOTE TESTNAME DQUOTE
# TESTLINE = INPUTQUOTED *WSP OUTPUTQUOTED *WSP TESTQUOTED GARBAGE
TESTPROGRAM=$1
TMPFILE=testharness.$$
trap "rm $TMPFILE" EXIT
N=0 # line number in the test file
while read -r line; do
N=$((N+1))
fields=$(perl -e 'split("\"",$ARGV[0]);print "$#_";' "$line")
if [[ $fields -lt 5 ]]; then
echo "skipped@$N"
continue
fi
INPUT=$(perl -e 'split("\"",$ARGV[0]);print "$_[1]";' "$line")
OUTPUT=$(perl -e 'split("\"",$ARGV[0]);print "$_[3]";' "$line")
TESTNAME=$(perl -e 'split("\"",$ARGV[0]);print "$_[5]";' "$line")
printf -- "$INPUT" | $TESTPROGRAM > $TMPFILE
perl -e "\$t='^\\\s*$OUTPUT\\\s*\$'; exit (<> =~ \$t);" < $TMPFILE
if [[ $? -ne 0 ]]; then # perl -e "exit(0==0)" => 1
echo "ok $TESTNAME"
else
echo -n "failed@$N $TESTNAME," \
"given: \"$INPUT\" expected: \"$OUTPUT\" received: "
cat $TMPFILE
echo
fi
done
परीक्षण मामलों का एक छोटा सा सेट:
"0 0" "0" "simple, 0+0=0"
"1 1" "2" "simple, 1+1=2"
"" "E" "error, no numbers"
"0" "E" "error, one number"
"0 0" "E" "error, two/too much white space"
"0 0 " "E" "error, trailing characters"
"01 0" "E" "error, leading zeros not allowed"
"-0 0" "E" "error, negative zero not allowed
"0 -0" "E" "error, negative zero not allowed here either
"1 1\n" "2" "LF only allowed trailing character
"\0001 1\n" "E" "error, try to confuse C string routines #1"
"1\00 1\n" "E" "error, try to confuse C string routines #2"
"1 \0001\n" "E" "error, try to confuse C string routines #3"
"1 1\000\n" "E" "error, try to confuse C string routines #4"
"1 1\n\000" "E" "error, try to confuse C string routines #5"
"-1 -1" "-2" "add all combinations of -1..1 #1"
"-1 0" "-1" "add all combinations of -1..1 #2"
"-1 1" "0" "add all combinations of -1..1 #3"
"0 -1" "-1" "add all combinations of -1..1 #4"
"0 0" "0" "add all combinations of -1..1 #5"
"0 1" "1" "add all combinations of -1..1 #6"
"1 -1" "0" "add all combinations of -1..1 #7"
"1 0" "1" "add all combinations of -1..1 #8"
"1 1" "2" "add all combinations of -1..1 #9"
"0 123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789" "123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789" "0+99 digits should work"
"100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "99 digits+99 digits should work"
"500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "test for accumulator overflow"
"-123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789 0" "-123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789" "0+negative 99 digits work"
"-100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 -100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "-200000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "99 digits+99 digits (both negative) should work"
"-500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 -500000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "-1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000" "test for negative accumulator overflow"
"1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890 0" "E" "error, 100 digits"