What is an "undefined expression?"?
114294
"undefined expression" means "has no mathematical meaning."\n\nYou don't even need two variables, you can use one e.g.:\n\nLim    (-1)^n\n n->inf\n\nLim    tan(alpha)\n alpha->90deg
674903
good answers, but there can also be multiple answers instead of just yes and no.\n\nfor instance, what day is today?\n\nif (day == monday)\n   cout<<"monday";\nelse if (day == tuesday)\n   cout<<"tuesday";\nelse if (day == wedensday)\n   cout<<"wedensday";\n...\nelse \n   cout<<"invalid entry";\n\nof course you could also use a case statement in this example, and would work the same.
393251
I'd dump it into a compiler and do it for you but you left too much out.  Impossible to return a syntax error free version without the whole program.\n\nAnyways, I get paid for this.  How many points is this worth?  ...to do someone's homework for him?  Or maybe he's doing contract work!  He's getting $70 and hour to let us do it for him!
557543
#include<iostream>\nusing namespace std;\nint main ()\n\n\n    {\nint number;\nint answer;\n    cout<<"Please enter number: ";\n    cin>>number;\n    \n    cout<<number<<"! = ";\n    answer = number;\n    while (number>1)\n\n\n        {\n        answer=answer*(number-1);\n        number--; \n    } \n\n    cout<<answer; \n    return 0; \n}
948798
It may not be with this line, the error may be from a line above, or so.  It would be easier to answer if you posted both the entire code you have, plus the error message you are receiving.  This way we can check it out.\n\nHint: Post the code and error message as Additional Details for this question.  Have a look at the green bar under your question, at the end you should see an icon that looks like a pen. Hover with your mouse over this, and look down the list that appears.  An item in that list should say 'Add Additional Details'.  Click on this.
37699
Syntax errors are errors that occur due to a mistyped function or command in programming languages. This usually happens due to a typo made in the function itself (they're pretty particular about how you spell certain things), or a mistake made in the format of the source code (for instance, in C++, you can run into a syntax error for leaving out an essential semi-colon).
805034
#include<iostream.h>\nvoid main()\n{int a,b,c,d,e;\ncout<<"Enter any two numbers&#92;n";\ncin>>a>>b;\nc=a+b;\nd=a*b;\ne=a/b;\ncout<<"&#92;nThe sum of <<a<<|" and "<<b<<" is:"<<c;\ncout<<"&#92;nThe product of <<a<<|" and "<<b<<" is:"<<d;\ncout<<"&#92;nThe qoutient of <<a<<|" and "<<b<<" is:"<<e;\ngetch();\n}
810128
Neither.\n0/0 is indeterminate, or undefined.
838865
What language is this?  It looks like the result of a night of unbridled passion between C, C++, and Java!\n\nSo, you know that "x->y" is equivalent to "(*x).y", right? So, even though you never used . directly, it's apparently what the compiler is seeing. Of course, . can only be used with a struct, and it is complaining that you are using it with something that is not a struct. However, the error is confusing, since the problem is really caused by a problem with pointers.\n\n Consider b[10]  (and b[9]).  b is a pointer to blocks. The [ ] operator dereferences this pointer, so that b[10] has type block, not block *.  Since -> can only be used with pointers (block *), the compiler generates an error. \n\nThink carefully about what you want to do, draw a picture on paper, and make sure your types all make sense.  Good luck.
1118582
What are you doing when you get this error?
127425
It means that whatver value you are testing for in your while clause has not yet been initialized or cannot be evaluated.
367158
(0! + 0! + 0! + 0! + 0!)!\n\n\n0^0 is undefined.
665768
if I understand your question, you are asking if the operators (and, or, not, +, -, etc) are a built in function like say printf or scanf. If that is what you are asking then, no, they arent. When you compile your code, the compiler converts them directly to an opcode in machine language, where printf, scanf, are library functions.
477862
I can see by looking at it that it won't compile.\nYou don't have semicolons at the end of the lines.\n\nI you put in the semicolons and replace the comma in the for loop with the another semicolon, it will compile and run.\n\nHowever, I don't think that you will get what you expect.\n\nYou use x without initializing it. That means that x will take the value that is already in memeory at that address.\n\nNext you condition your while loop on (R!=R) which is always false, so the while loop never executes.\n\nWhat you will print out is the last value that was left in x which is a location on the stack. \nI'm running MS C++ in debug mode, so it fills in the bytes of x with the value CC so you can recognize it as uninitialized stack memory.\n\nLast, you compute S but never use it.
142433
you havn't closed "void determinecost (...)" function with curly braces " } "\n\nin c++, you have to close any block of code with } that has been opened with {
338547
Seems like you have a type-o just above line 13.  in your declaration of  the Matrix class.  should be a single c in the word 'class'.\n\nIf you aren't already, you may want to consider using an IDE like netbeans which will catch simple  syntax errors like this for you before compiling.\n\nhope this helps.
971943
A variable expression is simply an expression -- a colllection of values and operators -- that contains at least one "variable" -- which is a quantity whose value has not yet been determined.\n\nFor example:\n\n1 + 9/3 is an expression.  It can be simplified to a single value, 4.\n\n1 + x is a variable expression.  It cannot be simplified to a single value, because 'x' represents an unknown quantity, a quantity that can vary.