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#include <iostream>
using namespace std;
int main()
{
cout.precision(50);
cout<<"Anmol's Square Root Calculator!"<<endl;
cout<<""<<endl;
cout << "This program will compute the square root\n";
cout << "of a number using the Babylonian algorithm!\n";
cout<<""<<endl;
cout<<"Only positive numbers work with this algorithm!"<<endl;
cout<<""<<endl;
cout<<"All cycle #1 approximate square roots are guessed\n"<<"using 1 as the first approximate."<<endl;
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
char playAgain='y';
while (playAgain !='n')
{
int count(25), cycle(1);
double guess, sqrt, num;
cout << "Please enter the number you would like to compute the square root of: ";
cin >> num;
cout<<""<<endl;
do {
if (num <= 0)
{
cout<<""<<endl;
cout<<"Invalid input. Please re-enter a valid number: ";
cin >> num;
cout<<""<<endl;
}
} while (num <= 0);
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
for (guess=1; count!=0; count--)
{
guess =(guess + (num/guess))/2;
cout<<""<<endl;
cout<<"Cycle "<<cycle<<" Aproximate: "<<guess<<endl;
sqrt = guess;
cycle++;
}
cout<<""<<endl;
cout << "The square root of "<< num << " is " << sqrt<<"!"<<endl;;
cout <<""<< endl;
do {
cout <<""<<endl;
cout << "Would you like to calculate again? (y/n): ";
cin >> playAgain;
cout <<" "<<endl;
do {
cout <<""<<endl;
if ((playAgain !='y' && playAgain !='n'))
{
cout << "Invalid input. Only y/n: ";
cin >> playAgain;
cout <<" "<<endl;
}
} while (playAgain !='y' && playAgain !='n');
} while (playAgain !='y' && playAgain !='n');
}
cout<<"Thank you for using a program made by Anmol Sethi!"<<endl;
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
cin.get();
return 0;
}
#include <iostream>
using namespace std;
int main()
{
cout.precision(18);
cout<<"Anmol's Square Root Calculator!"<<endl;
cout<<""<<endl;
cout << "This program will compute the square root\n";
cout << "of a number using the Babylonian algorithm!\n";
cout<<""<<endl;
cout<<"Only positive numbers work with this algorithm!"<<endl;
cout<<""<<endl;
cout<<"All cycle #1 approximate square are guessed\n"<<"using 1 as the approximate."<<endl;
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
char playAgain='y';
while (playAgain !='n')
{
int count(25), cycle(1);
double guess(1), sqrt, x, num;
cout << "Please enter the number you would like to know the square root of: ";
cin >> num;
cout<<""<<endl;
do {
if (num <= 0)
{
cout<<""<<endl;
cout<<"Invalid input. Please re-enter a valid number: ";
cin >> num;
cout<<""<<endl;
}
} while (num <= 0);
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
while (count > 0)
{
x = guess =(guess + (num/guess))/2;
cout<<""<<endl;
cout<<"Cycle "<<cycle<<" Aproximate: "<<guess<<endl;
sqrt = guess;
count-=1;
cycle+=1;
}
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
cout << "The sqaure root of "<< num << " is " << sqrt<<"!"<<endl;;
cout <<""<< endl;
do {
cout <<""<<endl;
cout << "Would you like to calculate again? (y/n): ";
cin >> playAgain;
cout <<" "<<endl;
do {
cout <<""<<endl;
if ((playAgain !='y' && playAgain !='n'))
{
cout << "Invalid input. Only y/n: ";
cin >> playAgain;
cout <<" "<<endl;
}
} while (playAgain !='y' && playAgain !='n');
} while (playAgain !='y' && playAgain !='n');
}
cout<<"Thank you for using a program made by Anmol Sethi!"<<endl;
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
cout<<""<<endl;
cin.get();
return 0;
}
最佳答案
首先,欢迎使用StackOverflow!我相信您在学习编程时会发现它是宝贵的资源。也就是说,要充分利用该网站,务必遵守the site's guidelines。通常,请记住,SO不是论坛而是Q&A站点,您会做的很好。
一般问题
您发布的代码存在一些问题:
A.空格
对于其他程序员和您自己而言,使用whitespace对making your code legible and intelligible都是非常有用的。 Tabbing to indicate scope(写作
bool condition = ...
if (condition) {
action();
}
bool condition = ...
if (condition) {
action()
}
cout
1.
只需编写
cout << "" << endl
,而不是编写
cout << endl
。
2.
有
differences between endl
and "\n"
,并且从您的代码中可能会发现它们。尽管使用两者的性能特征并不相同,但考虑到在此阶段肯定是
premature optimization。对于这种记录,坚持使用
endl
更具可读性。而不是
cout << "All cycle #1 approximate square are guessed\n" << "using 1 as the approximate." << endl;
cout << "All cycle #1 approximate square are guessed" << endl << "using 1 as the approximate." << endl;
3.
如果您的
inserting int
cout
远远超出一行代码(此代码,
cout << "This line of code is longer than feels comfortable." << endl << "It would be much more pleasant to break this into several lines so that I don't need to scroll to the right to view the whole thing." << endl;
cout << "Anmol's Square Root Calculator!" << endl;
cout << endl;
cout << "This program will compute the square root\n";
cout << "of a number using the Babylonian algorithm!\n";
cout << endl;
cout << "Only positive numbers work with this algorithm!" << endl;
cout << endl;
cout << "All cycle #1 approximate square are guessed\n" << "using 1 as the approximate." << endl;
cout << endl;
cout << endl;
cout << endl;
cout
上的插入运算符,您可以编写
cout << "Anmol's Square Root Calculator!" << endl
<< endl
<< "This program will compute the square root" << endl
<< "of a number using the Babylonian algorithm!" << endl
<< endl
<< "Only positive numbers work with this algorithm!" << endl
<< endl
<< "All cycle #1 approximate square are guessed" << endl
<< "using 1 as the approximate." << endl
<< endl
<< endl
<< endl;
main()
中。即使使用如此短的程序,仍然会影响可读性。更重要的是,这意味着您无法在其他地方重复使用此程序中创建的功能。
show a greeting
do {
get number from user
compute the squareroot of that number while logging
show the user the squareroot
ask the user if they want to do another computation
} while the user wants the program to keep running
show a farewell
cout
即可:
void showGreeting()
{
cout << "Anmol's Square Root Calculator!" << endl
<< endl
<< "This program will compute the square root" << endl
<< "of a number using the Babylonian algorithm!" << endl
<< endl
<< "Only positive numbers work with this algorithm!" << endl
<< endl
<< "All cycle #1 approximate square are guessed" << endl
<< "using 1 as the approximate." << endl
<< endl
<< endl
<< endl;
}
void showFarewell()
{
cout << "Thank you for using a program made by Anmol Sethi!" << endl
<< endl
<< endl
<< endl;
cin.get();
}
double getInputNumber()
{
double num = 0.0f;
cout << "Please enter the number you would like to compute the square root of: ";
cin >> num;
cout << endl;
do {
if (num <= 0)
{
cout << endl
<< "Invalid input. Please re-enter a valid number: ";
cin >> num;
cout << endl;
}
} while (num <= 0);
cout << endl
<< endl
<< endl;
return num;
}
main
的角度来看),这是一个非常布尔的问题:我们应该调用某个函数,并且应该返回
true
或
false
。但是,在该功能内,我们正在与用户进行交互,因此使用
'y'
和
'n'
是理想的选择(至少对于说英语的用户而言);
bool getRunAgain()
{
bool choice = false;
char playAgain = 'n';
do {
cout << endl
<< "Would you like to calculate again? (y/n): ";
cin >> playAgain;
cout << endl;
do {
cout << endl;
if ((playAgain !='y' && playAgain !='n'))
{
cout << "Invalid input. Only y/n: ";
cin >> playAgain;
cout << endl;
}
} while (playAgain !='y' && playAgain !='n');
} while (playAgain !='y' && playAgain !='n');
if (playAgain == 'y') {
choice = true;
} else /*if (playAgain == 'n')*/ {//At this, playAgain is either 'y' or 'n'. So if it's not 'y', it's 'n'.
choice = false;
}
return choice;
}
do...while
循环。其重复条件是
playAgain !='y' && playAgain !='n'
。乍一看,这似乎是有道理的:我们需要确保用户输入的是“是”或“否”。但是请注意,内部
do...while
循环的条件完全相同。这意味着除非
playAgain
等于
'y'
或
'n'
,否则内部循环不会退出。因此,当我们离开内部循环时,可以确定
playAgain
是
'y'
还是
'n'
。因此,我们无需再次检查。这使我们可以像下面这样重写函数:
bool getRunAgain()
{
bool choice = false;
char playAgain = 'n';
cout << endl
<< "Would you like to calculate again? (y/n): ";
cin >> playAgain;
cout << endl;
do {
cout << endl;
if ((playAgain !='y' && playAgain !='n'))
{
cout << "Invalid input. Only y/n: ";
cin >> playAgain;
cout << endl;
}
} while (playAgain !='y' && playAgain !='n');
if (playAgain == 'y') {
choice = true;
} else /*if (playAgain == 'n')*/ {
choice = false;
}
return choice;
}
do {
cout << endl;
if ((playAgain !='y' && playAgain !='n'))
{
cout << "Invalid input. Only y/n: ";
cin >> playAgain;
cout << endl;
}
} while (playAgain !='y' && playAgain !='n');
cout << endl
不太重要之外,除非
(playAgain !='y' && playAgain !='n')
,否则我们根本不想进入循环。这需要
while
循环而不是
do...while
循环:
while (playAgain !='y' && playAgain !='n')
{
cout << "Invalid input. Only y/n: ";
cin >> playAgain;
cout << endl;
}
bool getRunAgain()
{
bool choice = false;
char playAgain = 'n';
cout << endl
<< "Would you like to calculate again? (y/n): ";
cin >> playAgain;
cout << endl;
while (playAgain !='y' && playAgain !='n')
{
cout << "Invalid input. Only y/n: ";
cin >> playAgain;
cout << endl;
}
if (playAgain == 'y') {
choice = true;
} else /*if (playAgain == 'n')*/ {
choice = false;
}
return choice;
}
int count(25), cycle(1);
double guess, sqrt, num;
cout << "Please enter the number you would like to compute the square root of: ";
cin >> num;
cout << endl;
for (guess=1; count!=0; count--)
{
guess =(guess + (num/guess))/2;
cout<<""<<endl;
cout<<"Cycle "<<cycle<<" Aproximate: "<<guess<<endl;
sqrt = guess;
cycle++;
}
double computeSquareRootBabylonian(double num)
{
int count(25), cycle(1);
double guess, sqrt;
for (guess = 1; count != 0; count--)
{
guess = (guess + (num/guess))/2;
cout << endl
<< "Cycle " << cycle << " Aproximate: " << guess << endl;
sqrt = guess;
cycle++;
}
return sqrt;
}
1.
初始近似值(您正在使用的初始猜测
1
)实际上不是此函数的一部分。调用此函数的代码应指定起始点:
double computeSquareRootBabylonian
(double num,
double initialApproximation)
{
double approximation, sqrt;
unsigned int count(25), cycle(1);
for (approximation = initialApproximation; count != 0; count--)
{
approximation =(approximation + (num/approximation))/2;
cout << endl
<< "Cycle " << cycle << " Aproximate: " << approximation << endl;
sqrt = approximation;
cycle++;
}
return sqrt;
}
2.
迭代次数(或循环,正如您所称的)实际上也不是函数的一部分。调用此函数的代码必须指定该数字:
double computeSquareRootBabylonian
(double num,
double initialApproximation,
unsigned int iterations)
{
double approximation, sqrt;
unsigned int iterationsRemaining = iterations;
unsigned int iteration = 1;
for (approximation = initialApproximation; iterationsRemaining != 0; iterationsRemaining--)
{
approximation =(approximation + (num/approximation))/2;
cout << endl
<< "Cycle " << iteration << " Aproximate: " << approximation << endl;
sqrt = approximation;
iteration++;
}
return sqrt;
}
3.
这个
for
循环很奇怪。奇怪是因为您在初始化时初始化了变量
guess
(我已将其重命名为
approximation
),尽管它不是
"loop variable"。在第二段代码中,您使用了
while (count > 0)
{
guess =(guess + (num/guess))/2;
cout << endl;
cout << "Cycle " << cycle << " Aproximate: " << guess << endl;
sqrt = guess;
count-=1;
cycle+=1;
}
for
循环和
standard way可以使意图最清晰(在初始化中初始化相同的变量,循环的条件取决于该变量,并在每个循环的末尾进行更改)循环:
for (int i = 0; i < 10; i++)
)。在这里被迭代的变量是
count
(我称为
iterationsRemaining
),并且只要用户指定的迭代次数(
iterations
参数)被递减了
1
即可。它大于
0
。这种情况需要以下for循环:
for (unsigned int iterationsRemaining = iterations;
iterationsRemaining > 0;
iterationsRemaining--)
{
//...
}
approximation
初始化为
initialApproximation
:
double computeSquareRootBabylonian
(double num,
double initialApproximation,
unsigned int iterations)
{
double approximation = initialApproximation;
unsigned int iteration = 1;
double sqrt;
for (unsigned int iterationsRemaining = iterations;
iterationsRemaining > 0;
iterationsRemaining--)
{
approximation =(approximation + (num/approximation))/2;
cout << endl
<< "Cycle " << iteration << " Aproximate: " << approximation << endl;
sqrt = approximation;
iteration++;
}
return sqrt;
}
for
循环的好理由。您无需(在合理的范围内)更改代码以使其更快,除非您确定它的运行速度太慢。编写在阅读时有意义的代码更为重要。
4.
变量
sqrt
的作用是什么?它反复分配了
approximation
的值并从函数中返回,但返回时的值与
approximation
的值相同。因此,将其删除意味着过早的优化。但是,调用变量
sqrt
表示它是真正的平方根,而不是近似值。因此,应将其删除,让我们
double computeSquareRootBabylonian
(double num,
double initialApproximation,
unsigned int iterations)
{
double approximation = initialApproximation;
unsigned int iteration = 1;
for (unsigned int iterationsRemaining = iterations;
iterationsRemaining > 0;
iterationsRemaining--)
{
approximation =(approximation + (num/approximation))/2;
cout << endl
<< "Cycle " << iteration << " Aproximate: " << approximation << endl;
iteration++;
}
return approximation;
}
5.
在每次迭代中记录逐次逼近并不是您总是希望平方根函数执行的事情。我们应该添加一个参数,允许调用代码指定是否记录近似值:
double computeSquareRootBabylonian
(double num,
double initialApproximation,
unsigned int iterations,
bool log)
{
double approximation = initialApproximation;
unsigned int iteration = 1;
for (unsigned int iterationsRemaining = iterations;
iterationsRemaining > 0;
iterationsRemaining--)
{
approximation =(approximation + (num/approximation))/2;
if (log)
{
cout << endl << "Cycle " << iteration << " Aproximate: " << approximation << endl;
}
iteration++;
}
return approximation;
}
show a greeting
do {
get number from user
compute the squareroot of that number while logging
show the user the squareroot
ask the user if they want to do another computation
} while the user wants the program to keep running
show a farewell
int main(int argc, char *argv[])
{
//Do this stuff.
}
showGreeting()
和
showFarewell()
表示问候和告别:
int main(int argc, char *argv[])
{
showGreeting();
//do {
//get number from user
//compute the squareroot of that number while logging
//show the user the squareroot
//ask the user if they want to do another computation
//} while the user wants the program to keep running
showFarewell();
}
bool
,它表示用户是否要使用函数
getRunAgain()
计算另一个平方根。我们应该
do
用户输入和计算
while
getRunAgain()
!
int main(int argc, char *argv[])
{
showGreeting();
do {
//get number from user
//compute the squareroot of that number while logging
//show the user the squareroot
} while(getRunAgain());
showFarewell();
}
getInputNumber()
,它返回一个
double
,代表用户要计算其平方根的数字。我们需要使用两次此
double
,一次用作平方根函数
computeSquareRootBabylonian()
的参数,一次将输入及其平方根输出给用户。结果,我们将需要一个局部变量:
int main(int argc, char *argv[])
{
showGreeting();
do {
double number = getInputNumber();
//compute the squareroot of that number while logging
//show the user the squareroot
} while(getRunAgain());
showFarewell();
}
computeSquareRootBabylonian()
具有四个参数:
number
作为第一个参数。现在,我们可以
hardcode其余参数,使用
1
进行初始近似,使用
25
进行迭代,使用
true
进行是否记录。
computeSquareRootBabylonian()
的结果(注销结果时),因此无需使用局部变量就可以摆脱困境。为了清楚起见,让我们继续使用以下一种方法:
int main(int argc, char *argv[])
{
showGreeting();
do {
double number = getInputNumber();
double squareRoot = computeSquareRootBabylonian(number, 1.0f, 25, true);
//show the user the squareroot
} while(getRunAgain());
showFarewell();
}
int main(int argc, char *argv[])
{
showGreeting();
do {
double number = getInputNumber();
double squareRoot = computeSquareRootBabylonian(number, 1.0f, 25, true);
cout << endl
<< "The square root of " << num << " is " << sqrt << "!" << endl
<< endl;
} while(getRunAgain());
showFarewell();
}
cout
的精度。
int main(int argc, char *argv[])
{
cout.precision(50);
showGreeting();
do {
double number = getInputNumber();
double squareRoot = computeSquareRootBabylonian(number, 1.0f, 25, true);
cout << endl
<< "The square root of " << num << " is " << sqrt << "!" << endl
<< endl;
} while(getRunAgain());
showFarewell();
}
cout
的问题:
main
函数),我将小数的精度特别设置为15。但是,它输出的数字为12个小数。但是在此代码[第二个
main
函数]中,它以15位小数输出。 [...]我很困惑为什么会这样。我尝试了cout <<固定<< showpoint;但是当num无缘无故地输出15个零时,它看起来不愉快。谁能启发我为什么会这样以及如何解决?
operator <<
)转换为
std::cout
(通常转换为
std::ostream
)有关。它与使用
double
格式化
std::ostream::precision
输出有关。在将问题减少到这一点之后,我将开始查看相关文档,如果没有提供足够的解释,请使用这些关键字在StackOverflow上进行搜索。如果找不到解决方案,我将生成一个
example来说明我遇到的困难。
double
初始化为一个值,将
cout
的精度设置为
15
并将我们的
double
插入到
cout
中:
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
double value = 1.1111111111111111111111111f;
cout.precision(15);
cout << value << endl;
return 0;
}
1.11111111111111
1
。尽管这不是我们所希望的(小数点后是14位而不是12位),但看起来我们走在正确的轨道上。看起来正在发生某种舍入行为。也许我们应该将零放在小数点后的第13位和第14位:
#include <iostream>
using namespace std;
int main(int argc, char *argv[])
{
double value = 1.1111111111110011111111111f;
cout.precision(15);
cout << value << endl;
return 0;
}
1.111111111111
std::cout
输出
double
。我想显示小数点后的前15位数字,或者至少显示非零数字。但是以这段代码为例
int main(int argc, char *argv[])
{
double value = 1.1111111111110011111111111f;
cout.precision(15);
cout << value << endl;
return 0;
}
1.111111111111
而不是我期望的
1.111111111111001
。我可以使用
std::fixed
解决此问题
int main(int argc, char *argv[])
{
double value = 1.1111111111110011111111111f;
cout.precision(15);
cout << std::fixed << value << endl;
return 0;
}
1.111111111111001
。不幸的是,即使所有数字均为零,使用
std::fixed
也会始终在小数点后打印15位数字。例如来自的输出
int main(int argc, char *argv[])
{
double value = 1.0f;
cout.precision(15);
cout << std::fixed << value << endl;
return 0;
}
1.000000000000000
而不是我希望的
1
。
cout
输出最多15个十进制数字和更少的十进制数字?
cout
的精度设置为高于15多少,我们的输出中都只会获得15个十进制数字。让我们编写一个程序,该程序将
double
初始化为一个值,将
cout
的精度设置为
30
并将我们的
double
插入到
cout
中:
int main(int argc, char *argv[])
{
double value = 1.55f;
cout.precision(30);
cout << value << endl;
return 0;
}
1.5499999523162841796875
,该数字由小数点后的22位数字组成。这表明我们很不幸运地获得了准确的15位输出:我们遇到第一个问题时遇到的同种舍入行为显然是与平方根函数返回的值的输出进行交互,使得在这种舍入行为下,小数点后第十五位被打印。
double
中插入一个
cout
,以便显示尽可能多的非零十进制数字,最多15个。
double
中
cout
的字符串表示形式应至少满足以下测试:
x.xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxf ----> "x.xxxxxxxxxxxxxxx"
x.xxxxx0000000000000000000000000f ----> "x.xxxxx"
x.000000000000000000000000000000f ----> "x"
x.00000000000000x000000000000000f ----> "x.00000000000000x"
std::ostream::precision
不能以任何明显的方式实现。正如您所指出的,我们不能使用
std::fixed
,因为这肯定会使上面的第三项测试失败。我们也不能轻易使用默认值,因为如
cplusplus.com所述,
double
并返回
std::string
的函数:
std::string stringFromDouble(double value, unsigned int decimalDigits)
{
int integerComponent = abs((value > 0) ? (int)(floor(value)) : (int)(ceil(value)));
std::ostringstream integerStream;
integerStream << integerComponent;
std::string integerString = integerStream.str();
unsigned int integerLength = integerString.length();
std::ostringstream stream;
stream.precision(decimalDigits + integerLength);
stream << value;
std::string str = stream.str();
return str;
}
关于c++ - 设置精度无法正常工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14290662/
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