blocks|key|2338290|text|您至少有三个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2338291|2338292|表达式5/9*(t-32)的求值顺序是从左到右，因此5/9部分是一个整数(截断)除法，始终生成0。零点与任何有限点的乘积是零。有几种方法可以解决这个问题，但更简单的方法之一是重写表达式，因为5+*+(t+-+32)+/+9.|ordered-list-item|offset|length|style|CODE|2338293|Your变量t和celcius是标量。他们每个人一次只持有一个号码。在您的第一个循环中，您将多个值按顺序分配给每个循环。当您稍后执行第二次循环以打印结果时，只能访问分配给每个变量的最后一个值。如果您必须将输出推迟到读取所有输入之后，那么处理它的一种方法是使t和celcius数组具有足够的大小，并将值存储在不同的元素中。(还要注意:英语单词的正确拼写是"celsius".)|2338294|Per+@albert+in+comments，在索引的do循环结束后，迭代变量的值是它在下一次循环迭代中的值(如果有的话)。因此，通过将变量n同时用作迭代变量和上限，可以使每次循环后的值与以前不同。有几种方法可以解决这个问题，但我强烈建议您简单地避免重用n作为迭代变量。避免有目的的迭代变量并不能提高效率。|2338295|2338296|entityMap^0|0|0|3|A|Q|3|2N|G|0|6|1|8|7|3K|1|3M|7|0|S|2|1Z|1|3K|1|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|R|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|S|8|@$F|T|G|U|H|I]|$F|V|G|W|H|I]|$F|X|G|Y|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|Z|8|@$F|10|G|11|H|I]|$F|12|G|13|H|I]|$F|14|G|15|H|I]|$F|16|G|17|H|I]]|9|@]|A|$]]|$1|L|3|M|5|E|7|18|8|@$F|19|G|1A|H|I]|$F|1B|G|1C|H|I]|$F|1D|G|1E|H|I]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|P|$]]

You have at least three problems:

<ol>
<li>The expression <code>5/9*(t-32)</code> is evaluated left-to-right, so the <code>5/9</code> part is an integer (truncating) division, always producing 0. The product of zero with anything finite is zero. There are several ways you could address that, but one of the simpler would be to rewrite the expressions as <code>5 * (t - 32) / 9</code>.</li>
<li>Your variables <code>t</code> and <code>celcius</code> are scalars. They hold only one number at a time each. In your first loop you assign multiple values to each of them them in sequence. When you later perform a second loop to print the results, only the last value assigned to each variable will be accessible. If you must defer output until after reading all the input then one way to handle it would be to make <code>t</code> and <code>celcius</code> arrays of sufficient size, and store your values in different elements. (Also note: the correct spelling of the English word is "celsius".)</li>
<li>Per @albert in comments, after an indexed <code>do</code> loop finishes, the value of the iteration variable is the one it would have had in the next iteration of the loop, if there were one. Therefore, by using variable <code>n</code> both as your iteration variable and your upper bound, you are causing its value to be different after each loop than it was before. There are several ways you could address this, but I urge you to simply avoid reusing <code>n</code> as your iteration variable. There is no efficiency to be gained by avoiding a for-purpose iteration variable.</li>
</ol>

blocks|key|2549806|text|您有几个选项。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2549807|2549808|程序温度！F+C温度转换隐式无实数，可分配:：华氏(：)整数：：n打印*，‘输入行数’读取*，n分配(华氏(N))打印*，‘输入t的值:一行上的所有项目’读取*，华氏打印*，'F：'，华氏打印*，'C:'，fahrenheit_to_celcius(华氏)包含纯元素函数fahrenheit_to_celcius(t_f)+fahrenheit_to_celcius(T_c)+real，|2549809|程序温度！F+C温度转换隐式无实数:：华氏整数：：i，n打印*，‘输入行数’读取*，n+do+i=+1，n打印*，‘输入t的值:每行一个'，n读取*，华氏打印*，'F：’，华氏，'C:'，fahrenheit_to_celcius(华氏)+enddo包含纯元素函数fahrenheit_to_celcius(t_f)+result(t_c)+real，intent(in)+:：t_f+real+:：t_c+t_c+=+5.*(t_f-32.)/9。end+function+fahrenheit_to_celcius+end+program|2549810|注意，我在函数中使用了elemental关键字。这意味着你既可以传递标量，也可以传递数组。对于直接计算来说，这是一个很好的解决方案:例程fahrenheit_to_celcius在这两种情况下都是相同的。|offset|length|style|CODE|2549811|我修复了5/9+(它返回0)和混合变量。|2549812|entityMap^0|0|0|0|0|1W|L|0|4|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|R|8|@]|9|@]|A|$]]|$1|C|3|D|5|6|7|S|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|P|$]]

You have several options.

<ol>
<li>Store the input in an array and process the array.

<pre><code>program temperature
 ! F C temperature conversion
 implicit none

 real, allocatable :: fahrenheit(:)
 integer:: n

 print *,'enter the number of lines'
 read*, n
 allocate(fahrenheit(n))
 print*,'enter the value of t: all items on one line'
 read*, fahrenheit
 print *, 'F: ', fahrenheit
 print *, 'C:', fahrenheit_to_celcius(fahrenheit)

contains

 pure elemental function fahrenheit_to_celcius(t_f) result(t_c)
 real, intent(in) :: t_f
 real :: t_c

 t_c = 5.*(t_f-32.)/9.

 end function fahrenheit_to_celcius

end program
</code></pre></li>
<li>Process the inputs one at a time

<pre><code>program temperature
 ! F C temperature conversion
 implicit none

 real :: fahrenheit
 integer:: i, n

 print *,'enter the number of lines'
 read*, n

 do i = 1, n
 print*,'enter the value of t: one per line',n
 read*, fahrenheit
 print *, 'F: ', fahrenheit, 'C:', fahrenheit_to_celcius(fahrenheit)
 enddo

contains

 pure elemental function fahrenheit_to_celcius(t_f) result(t_c)
 real, intent(in) :: t_f
 real :: t_c

 t_c = 5.*(t_f-32.)/9.

 end function fahrenheit_to_celcius

end program
</code></pre></li>
</ol>

Note that I have used the elemental keyword for the function. It means that you can pass a scalar as well as an array. This is a nice solution for direct computations such as the one here: the routine <code>fahrenheit_to_celcius</code> is identical in both cases.

I fixed the <code>5/9</code> (that returns 0) and the mixed up variables as well.

When I run the code for Celsius only I am getting the result below the code:

<pre><code> program temperature
! F C temperature conversion
implicit none
real :: celcius=0.0, fahrenheit=0.0

integer:: t,n
print *,'enter the number of lines'
read*, n
do n=1,n
print*,'enter the value of t: one per line',n
read*, t
celcius=5/9*(t-32)

enddo
do n=1,n
print*, t, celcius
enddo
end program
</code></pre>

result

<pre><code> enter the number of lines
3
 enter the value of t: one per line 1
50
 enter the value of t: one per line 2
20
 enter the value of t: one per line 3
10
 10 0.00000000E+00
 10 0.00000000E+00
 10 0.00000000E+00
 10 0.00000000E+00
</code></pre>

It's clear that compiler is not picking the value of <code>t</code> in the calculation.

Fortran do loop not calculating the entered variable

编译器

当我只运行摄氏度的代码时，我得到了下面的代码结果：    program temperature! F C temperature conversionimplicit nonereal :: celcius=0.0, fahrenheit=0.0integer:: t,nprint *,'enter the numb...

问Fortran do循环不计算输入的变量
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