长度大于10个字符的缩写代码，不能正常运行

这个缩写是这样做的:我们写下一个单词的第一个和最后一个字母，在这两个字母之间写下第一个和最后一个字母之间的字母数。该数字是十进制的，并且不包含任何前导零。

因此,"localization"将拼写为"l10n"one_answers"internationalization"将被拼写为"i18n"

这是我的代码:

#include<iostream>
#include<stdlib.h>
#include<string>
using namespace std;
int main()
{
int n,c;
std::string str;
c=0;
char word[n][100];
cin>>n;
for(int i=0;i<n;i++)
{
cin>>word[i];
}
for(int j=0;j<n;++j)
{
str = word[j];
c = str.length();
if(c>10)
cout<<str[0]<<c-2<<str.back();
else
cout<<word[j]<<"n";
c=0;
}
}

谁能告诉我我应该改变什么?

我们是来帮忙的。好了，任务明白了。现在回答你的问题:

谁能告诉我我应该改变什么?

首先，程序不能在我的机器上编译，因为不兼容的语言结构，如VLA(可变长度数组)。这不是合法的c++代码。然后有一个主要的错误，您使用未初始化的变量n来设置(也是禁止的)VLA的维度。这将导致未定义的行为或运行时崩溃。

顺便说一下，如果你为你的编译器启用了所有的警告，例如-wall -pedantic，那么编译器会为你找到所有的问题。让我们在这里列出更多的问题。

你不应该使用c函数，也不应该包含像" stdlib.h "这样的c头文件
不使用using namespace std;。始终使用完全限定名
在定义变量时总是初始化所有变量
使用有意义的、描述性的变量名，而不是像n、c这样的单一字母名称

std::array

std::vector

你知道std::string，因为你使用它。因此，请使用std::string字符串。永远不要使用旧的c风格字符串与char[]或char*
最后一条语句c=0;不需要。c将在下一轮被strlen覆盖。
对于单词的大小，您使用神奇的数字100。不应该使用魔术数字。如果用户输入的字符串超过100个字符。灾难将会发生。如果有的话，那么在这种情况下请使用cin.get。但最好是使用std::string。

好了，有了上面的智慧，让我们开始重构的第一步。我们将保留您的编程风格，但必须使用新的原始指针将VLA替换为动态内存管理。请注意:在c++中，New和raw指针不能用于自有内存。无论如何。第一步，我们将使用它。当然，对于字符串，我们将使用std::string。

#include<stdlib.h>
#include<string>
using namespace std;
int main()
{
int n,c;
std::string str;
c=0;
cin>>n;
std::string* word = new std::string[n];
for(int i=0;i<n;i++)
{
cin>>word[i];
}
for(int j=0;j<n;++j)
{
str = word[j];
c = str.length();
if(c>10)
cout<<str[0]<<c-2<<str.back() << "n";
else
cout<<word[j]<<"n";
c=0;
}
delete[] word;
}

下一步。删除不必要的东西。使用有意义的变量名和注释。

#include<iostream>
#include<string>
// Program to create abbreviations for words
// This abbreviation is made like this: we write down the first and the last letter
// of a word and between them we write the number of letters between the first and 
// the last letters. That number is in decimal system and doesn't contain any leading 
// zeroes.
// Thus, "localization" will be spelt as "l10n", and "internationalization" will be 
// spelt as "i18n".
// Words above this length will be transformed
constexpr unsigned int MaxWordLength = 10u;
int main()
{
// We will get the numberOfWords from the user and check for a valid input
unsigned int numberOfWords{};
if ((std::cin >> numberOfWords) and (numberOfWords > 0)) {

// Here we will store all words read from the user
std::string* words = new std::string[numberOfWords]{};

// Now read the specified amount of words from user
for(unsigned int index=0; index<numberOfWords; ++index)  {
std::cin >> words[index];
}

// Abbreviation activity
for(unsigned int index=0; index<numberOfWords; ++index) {

// Check length of word
const std::string& word = words[index];
const unsigned int wordLength = word.length();

// If word is longer than threshold, do abbreviation
if(wordLength > MaxWordLength)

// Print abbreviated word
std::cout << word[0] << wordLength-2 << word.back() << 'n';
else
// Print word as is
std::cout << word << 'n';
}
delete[] words;
}
else std::cerr << "nError: Invalid inputnn";
}

接下来，让我们摆脱new和使用std::vector。这不会影响整个程序。它是等价的。但是我们将使用更多的改进，比如基于范围的for循环。

#include <iostream>
#include <string>
#include <vector>
// Program to create abbreviations for words
// This abbreviation is made like this: we write down the first and the last letter
// of a word and between them we write the number of letters between the first and 
// the last letters. That number is in decimal system and doesn't contain any leading 
// zeroes.
// Thus, "localization" will be spelt as "l10n", and "internationalization" will be 
// spelt as "i18n".
// Words above this length will be transformed
constexpr size_t MaxWordLength = 10u;
int main()
{
// We will get the numberOfWords from the user and check for a valid input
size_t numberOfWords{};
if ((std::cin >> numberOfWords) and (numberOfWords > 0)) {

// Here we will store all words read from the user
std::vector<std::string> words(numberOfWords);

// Now read the specified amount of words from user
for(std::string& word : words)  {
std::cin >> word;
}

// Abbreviation activity
for(std::string& word : words) {

// Check length of word
const size_t wordLength = word.length();

// If word is longer than threshold, do abbreviation
if(wordLength > MaxWordLength)

// Print abbreviated word
std::cout << word[0] << wordLength - 2u << word.back() << 'n';
else
// Print word as is
std::cout << word << 'n';
}
}
else std::cerr << "nError: Invalid inputnn";
}

现在已经很好了。

如果您想查看更高级的c++解决方案，您可以查看以下内容:

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <iterator>
// Program to create abbreviations for words
// This abbreviation is made like this: we write down the first and the last letter
// of a word and between them we write the number of letters between the first and 
// the last letters. That number is in decimal system and doesn't contain any leading 
// zeroes.
// Thus, "localization" will be spelt as "l10n", and "internationalization" will be 
// spelt as "i18n".
// Words above this length will be transformed
constexpr size_t MaxWordLength = 10u;
int main()
{
// We will get the numberOfWords from the user and check for a valid input
if (size_t numberOfWords; (std::cin >> numberOfWords) and (numberOfWords > 0)) {

// Here we will store all words read from the user
std::vector<std::string> words(numberOfWords);

// Now read the specified amount of words from user
std::copy_n(std::istream_iterator<std::string>(std::cin), numberOfWords, words.begin());

// Abbreviation activity and output
std::transform(words.begin(), words.end(), std::ostream_iterator<std::string>(std::cout,"n"), [](const std::string& w){
return (w.length() > MaxWordLength) ? (w[0]+std::to_string(w.length()-2)+w.back()) : w;});
}
else std::cerr << "nError: Invalid inputnn";
}

相关内容

最新更新

热门标签：