Question

这只是我作业的第一部分，我修复了所有其他编译错误，但我一直收到这个错误，有五个。

1>\takehome\main.cpp(39) : error C2065: 'j' : undeclared identifier
1>\takehome\main.cpp(44) : error C2065: 'j' : undeclared identifier
\takehome\main.cpp(45) : error C2065: 'j' : undeclared identifier
\takehome\main.cpp(76) : error C2065: 'j' : undeclared identifier
\takehome\main.cpp(80) : error C2065: 'j' : undeclared identifier

我已尝试用它做任何事情，但我可能做错了什么......显然我是。如果你不介意我可以使用一些帮助:)。顺便说一句，万一有人想知道，做简单的。

#include <iostream>
using namespace std;

int main()
{
 int memory[100]; //Making it 100, since simpletron contains a 100 word mem.

 int operation; //taking the rest of these variables straight out of the book seeing as how they were italisized.

 int operand;

 int accum = 0; // the special register is starting at 0

 int position = 0; //making the starting position to be 0.

 for ( int j = 0; j < 100; j++ ) //Simply stating that for int j is = to 0, j must be less than 100 because that is the memory limit, and for every pass-through, increment j.

  memory[j] = 0;


 // This is for part a, it will take in positive variables in a sent-controlled loop and compute + print their sum. These are random variables.
 memory [0] = 2942;

 memory [1] = 2342;

 memory [2] = 3523;

 memory [3] = 2031;

 memory [4] = 5000;

 memory [5] = 8080;

 memory [6] = 3425;

 j = 0; //Makes the variable j start at 0.

 while ( true )
 {

  memory[ j ]%100 = operand; // Finds the op codes from the limit on the memory (100)
  memory[ j ]%100 = operation;

  //using a switch loop to set up the loops for the cases
  switch ( operation ){
   case 1: //reads a variable into a word from loc.
    cout <<"\n Input a positive variable:  ";
    cin >> memory[ operand ]; break;

   case 2: // takes a word from location
    cout << "\n\nThe content at location " << operand << "is " << memory[operand]; break;

   case 3:// loads
    accum = memory[ operand ]; break;

   case 4: //stores
    memory[ operand ] = accum; break;

   case 5: //adds
    accum = accum + memory[ operand ]; break;


   case 6: // subtracts
    accum = accum - memory[ operand ]; break;

   case 7: //divides
    accum = accum / (memory[ operand ]); break;

   case 8: // multiplies
    accum = accum*memory [ operand ]; break;

   case 9: // Branches to location
    j = -1; break;

   case 10: //branches if acc. is < 0
    if (accum < 0)
    j = 5; break;

   case 11: //branches if acc = 0
    if (accum == 0); break;

   case 12: // Program ends
    exit(0); break;
 }
 j++;
 }
return 0;
}

Answer 1

在“for”循环之外声明“j”。当你在里面声明循环标题时，它在循环块的本地，在它之外是不可见的。

Answer 2

当您在for语句中声明变量时，该变量仅在for循环体的范围内，例如。

for ( int j = 0; j < 100; j++ ) 
{
  // j is defined in here
}

// But j is undefined out here

Answer 3

您设置的j = 0未声明为int j = 0。

你是在for循环中完成的，但它的局部范围仅适用于循环体..

Answer 4

当你有类似

的东西时

for ( int i = 0; i < k; ++i )
{ 
// stuff 
}

i仅在for循环的范围内可用，因此，如果您执行以下操作：

for ( int i = 0; i < k; ++i )
{ 
// stuff 
}
cout << i;

您在cout << i;收到编译错误，因为i循环结束后for不再存在。

Answer 5

在C ++中，for循环中声明的变量的范围是循环。所以当你说：

for ( int j = 0; j < 100; j++ ) {
    // j only exists here (and in the for statement itself)
}

变量j仅存在于循环体中。

Answer 6

变量j是for循环的本地变量。您需要通过执行以下操作来扩大其范围：

int j;
for(j = 0; j < 100; ++j)

或稍后重新宣布：

for(int j=0; j<100; ++j)
...
...
int j = 0;
while(true) 
...

Answer 7

j仅存在于for循环中，即在指令

中

for ( int j = 0; j < 100; j++ ) //Simply stating that for int j is = to 0, j must be less than 100 because that is the memory limit, and for every pass-through, increment j.
    memory[j] = 0;

你应该写

int j;
for(j=0;j<100;j++)
...

尝试做这个家庭作业，但我不断收到编译错误

7 个答案: