此程序的目标是读取一行字符，以计数和识别该行中的字数。它必须确定这些单词中有多少是MMDDYY格式的日期。
例如，如果用户输入："四分和死亡队长前"
输出应如下所示。
总字数：6
有效日期数：0
如果用户输入"四032356团队铁人七231290前"
输出应为
总字数：7
有效日期数：1
我在实现validateDate函数时遇到问题。我也不知道我的日期验证方法是否正确。如果代码一团糟，我最诚挚的道歉，我正在学习基础知识。提前谢谢。这是代码：
// #include "stdafx.h"
// This program demonstrates the use of dynamic variables
// Dean DeFino
#include <iostream>
#include <cctype>
#include <cstring>
using namespace std;
const int MAXNAME = 10;
//Function prototype
void processCstring(char *);
bool validateDate(char[], int);
int main()
{
int pos;
char *name = nullptr;
int result;
const int SIZE = 81; // The maximum size of the C-string.
char *cstring; // To hold a C-string
char goAgain; // To hold Y or N
cstring = new char[SIZE]; // Fill in code to allocate the character array pointed to by cstring
name = new char[SIZE]; // Fill in code to allocate the character array pointed to by name
cout << "Enter your last name with exactly 10 characters." << endl;
cout << "If your name has < 10 characters, repeat last letter. " << endl;
cout << "Blanks at the end do not count." << endl;
for (pos = 0; pos < MAXNAME; pos++)
cin >> *(name + pos); // Fill in code to read a character into the name array
// WITHOUT USING a bracketed subscript
cout << "Hi ";
for (pos = 0; pos < MAXNAME; pos++)
cout << *(name + pos);
do
{
cin.ignore();
// Get a C-string.
cout << "nEnter a C-string containing 80 or fewer characters:n";
cin.getline(cstring, SIZE);
// Display the number of words in the C-string.
cout << "nThe number of words in the C-string: ";
processCstring(cstring);
// Does the user want to do this again?
cout << "Process another string? (Y or N) ";
cin.get(goAgain);
// Validate the input. If the user enter something diferent of y,Y or n,N
//then ask again for an answer.
while ((toupper(goAgain) != 'Y') && (toupper(goAgain) != 'N'))
{
cout << "Please enter Y or N: ";
cin.get(goAgain);
}
} while (toupper(goAgain) == 'Y');
// Fill in code to deallocate cstring and name (two steps each pointer)
delete cstring;
cstring = nullptr;
delete name;
name = nullptr;
return 0;
}
//***************************************************
// Function processString *
// This function counts the number of words and the *
// number of valid datres in a string passed into an str. *
//***************************************************
void processCstring(char *cStrPtr)
{
char *str = cStrPtr;
int numWords = 0;
int numDates = 0;
int count = 0; // Variable para que el loop no sea infinito.
//****WARNING
//Do NOT RUN this while loop until you fixed.
//This is infinite loop
while (*(str + count) != ' ')
{
if (isspace(*(str + count))) //Complete the code to count the number of words.
count++; //Skip any leading spaces
else if (isalnum(*(str + count)) && isspace(*(str + count + 1)))
numWords++;
count++;
(validateDate(str, strlen(str));
cout << "h";
//Now count the number of valid dates in the form MMDDAA
//for example 040390 is a valid date, 230490 is not valid
// 23jun90 is not valid.
//Use the function validateDate (pass the word you found as
//parameter) to validate if the word is a valid date
}
cout << "nTotal number of words: " << numWords + 1 << endl;
cout << "Number of valid dates: " << numDates << endl;
}
//***************************************************
// Function validateDate *
// This function validate if a word received as a parameter *
// is a valid date. *
//***************************************************bool
validateDate(char myPass[], int size)
{
bool answer = false;
int i = 0; // Arreglar
while (i < size && !answer)
{
/*Validar que sea un string de 6 chars
Que el primer char1 < 2
char1 + char2 <= 9
char3 < 4
char3 + char4 <= 11*/
if (static_cast<int>(myPass[0]) < 2)
{
if (static_cast<int>(myPass[0]) + static_cast<int>(myPass[1]) <= 9)
{
if (static_cast<int>(myPass[2]) < 4)
{
if (static_cast<int>(myPass[2]) + static_cast<int>(myPass[3]) <= 11)
{
answer = true;
i++;
}
}
}
}
}
return answer;
}
对我来说，最简单的方法是使用正则表达式进行日期验证。你可以在这里阅读更多关于它们的信息。C中的正则表达式：示例？并在Inetnet中搜索regexec
还有一种C++方法：http://www.cplusplus.com/reference/regex/
validateDate（）的方法看起来不正确，也太复杂了。
我也不同意将regexp用于此目的的建议，因为它会使任务更加复杂。
我的版本想要：
// this method could be improved
int parseDD(const char * in) {
if (not std::isdigit(in[0]) or not std::isdigit(in[1])) {
return -1;
}
return (in[0] - '0') * 10 + (in[1] - '0');
}
bool validateDate(const char myPass[], int size) {
if (size != 6) {
return false;
}
static int daysInMonth[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int mm = parseDD(myPass);
int dd = parseDD(myPass+2);
int yy = parseDD(myPass+4);
if (yy < 0) {
return false;
}
if (not (mm > 0 and mm <= 12)) {
return false;
}
int days = daysInMonth[mm];
bool leap = yy % 4 == 0;
if (mm == 2 and leap) {
days += 1; // 29 Feb
}
if (not (dd > 0 and dd <= days)) {
return false;
}
return true;
}
伙计们，我用了一种不同的方法。感谢您的回复。我肯定知道我的静态广播有问题。
（这是代码）
#include <iostream>
#include <cctype>
#include <cstring>
using namespace std;
const int MAXNAME = 10;
//Function prototype
void processCstring(char *);
bool validateDate(char[], int);
int main()
{
int pos;
char *name = nullptr;
int result;
const int SIZE = 81; // The maximum size of the C-string.
char *cstring; // To hold a C-string
char goAgain; // To hold Y or N
cstring = new char[SIZE]; // Fill in code to allocate the character array pointed to by cstring
name = new char[SIZE]; // Fill in code to allocate the character array pointed to by name
cout << "Enter your last name with exactly 10 characters." << endl;
cout << "If your name has < 10 characters, repeat last letter. " << endl;
cout << "Blanks at the end do not count." << endl;
for (pos = 0; pos < MAXNAME; pos++)
cin >> *(name + pos); // Fill in code to read a character into the name array
// WITHOUT USING a bracketed subscript
cout << "Hi ";
for (pos = 0; pos < MAXNAME; pos++)
cout << *(name + pos);
do
{
cin.ignore();
// Get a C-string.
cout << "nEnter a C-string containing 80 or fewer characters:n";
cin.getline(cstring, SIZE);
// Display the number of words in the C-string.
cout << "nThe number of words in the C-string: ";
processCstring(cstring);
// Does the user want to do this again?
cout << "Process another string? (Y or N) ";
cin.get(goAgain);
// Validate the input. If the user enter something diferent of y,Y or n,N
//then ask again for an answer.
while ((toupper(goAgain) != 'Y') && (toupper(goAgain) != 'N'))
{
cout << "Please enter Y or N: ";
cin.get(goAgain);
}
} while (toupper(goAgain) == 'Y');
// Fill in code to deallocate cstring and name (two steps each pointer)
delete cstring;
cstring = nullptr;
delete name;
name = nullptr;
return 0;
}
//***************************************************
// Function processString *
// This function counts the number of words and the *
// number of valid datres in a string passed into an str. *
//***************************************************
void processCstring(char *cStrPtr)
{
char *str = cStrPtr;
int numWords = 0;
int numDates = 0;
int size = 0;
bool started = false;
//****WARNING
//Do NOT RUN this while loop until you fixed.
//This is infinite loop
while (*str != ' ')
{
if (started && isspace(*str))
{
if (validateDate(str - size, size))
{
numWords++;
numDates++;
}
else
numWords++;
size = 0;
}
else if (!isspace(*str))
{
if (started)
size++;
else
started = true;
}
str++;
}
if (size > 0)
{
if (validateDate(str - size, size))
numDates++;
else
numWords++;
}
//numDates = validateDate(str, strlen(str)); Test de la variable
cout << "nTotal number of words: " << numWords << endl;
cout << "Number of valid dates: " << numDates << endl;
}
//***************************************************
// Function validateDate *
// This function validate if a word received as a parameter *
// is a valid date. *
//***************************************************
bool validateDate(char myPass[], int size)
{
bool answer = false;
int i = 0;
while (i<size && !answer)
{
if (i == 0)
{
if (myPass[i] == '0')
i++;
else if (myPass[i] == '1')
{
if (myPass[i + 1] != '0' && myPass[i + 1] != '1' && myPass[i + 1] != '2')
i = 7;
else
i++;
}
else
i = 7;
}
if (i == 1)
i++;
if (i == 2)
{
if (myPass[i] == '0' || myPass[i] == '2')
i++;
else if (myPass[i] == '3')
{
if (myPass[i + 1] != '0' && myPass[i + 1] != '1')
i = 7;
else
i++;
}
else
i++;
}
if (i == 3)
{
if (isdigit(myPass[i]))
i++;
else
i = 7;
}
if (i == 4)
{
if (isdigit(myPass[i]))
answer = true;
else
i = 7;
}
}
return answer;
}