C++ Expression Must Have Pointer To Class Type ” Error In C++

Problem:

I've got a problem with c++ code. I am attaching the sample code snippet below. Please pull me up. Yee, of course, I am pretty new in C++

#include using namespace std;class Myclass{ public: void myfunc() { cout

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askedMar 7, 2020Gavin●12 ●7 ●7

Solution:

The code snippet you attached is triedto apply the .

Đang xem: Expression must have pointer to class type

operator on a pointer to an object, rather than on an object itself. That is the reason the code throws an “expression must have class type”​ error.

You should write this way:

#include using namespace std;class Myclass{ public: void myfunc() { coutI hope you get it.

answeredMar 7, 2020Naymulhasan●13 ●9 ●6

Your comment on this answer:

; The above codes throw an error expression must have a constant value c++. What’s wrong with it?

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askedApr 7, 2020Gavin15.3k points
0 votes
1 answer29 views
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c++ expression must be a modifiable l value
Problem: I have this following code: int M = 3; int C = 5; int match = 3; for ( int k =0; k

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askedDec 23, 2020Mashhoodch13k points
1 vote
1 answer78 views
78 views
c++ expression must be a modifiable value
Problem: I have a sample c++ code below that throws me back an error: “c++ expression must be a modifiable value” What could be the reason? int A = 3; int B = 5; int match = 3; for ( int k =0; k

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askedMar 10, 2020Gavin15.3k points
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Solution:

The “expressionshould have class type” is an error that is thrown at the timethe . operator, which is generallyappliedto access an object’s fields and methods, is used on pointers to objects.

At the time appliedon a pointer to an object, the . operator will attemptto tracethe pointer’s fields or systems, howeverit won’t, sincethey do not subsist. These fields or methods are part of an object, morethan a part of the pointer to an object.

Wrong code

The code below attemptsto imbedthe . operator on a pointer to an object, morethan on an object itself. The code throws an “expression shouldhave class type”​ error.

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using namespace std;class Myclass{ public: void myfunc() { cout

Correct code

The code below accuratelyembedsthe . operator on an object to access the object’s member functions. Becausea is now a class object, the code does not throw any error.

#include using namespace std;class Myclass{ public: void myfunc() { cout

answeredMay 28, 2020Tushar Shuvro●10 ●6 ●4
31,120 points
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