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Views: 650 | Replies: 15 | Solved
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Join Date: Jul 2008
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Light Poster
I am trying to create a program that gets the solution of a quadratic formula, but i keep getting an error and i cant figure it out.
1>quad.cpp(14) : error C2440: '=' : cannot convert from 'int (__cdecl *)(void)' to 'int'
1> There is no context in which this conversion is possible
1>quad.cpp(16) : error C2440: '=' : cannot convert from 'int (__cdecl *)(void)' to 'int'
1> There is no context in which this conversion is possible
1>quad.cpp(18) : error C2440: '=' : cannot convert from 'int (__cdecl *)(void)' to 'int'
1> There is no context in which this conversion is possible
1>quad - 3 error(s), 21 warning(s)
Here is the program
1>quad.cpp(14) : error C2440: '=' : cannot convert from 'int (__cdecl *)(void)' to 'int'
1> There is no context in which this conversion is possible
1>quad.cpp(16) : error C2440: '=' : cannot convert from 'int (__cdecl *)(void)' to 'int'
1> There is no context in which this conversion is possible
1>quad.cpp(18) : error C2440: '=' : cannot convert from 'int (__cdecl *)(void)' to 'int'
1> There is no context in which this conversion is possible
1>quad - 3 error(s), 21 warning(s)
Here is the program
#include <iostream>
#include <stdio.h>
#include "simpio.h"
#include "strlib.h"
#include "random.h"
#include "math.h"
using namespace std;
int main()
{
int a, b, c, x, y, z;
printf("Enter A: ");
a = GetInteger;
printf("Enter B: ");
b = GetInteger;
printf("Enter c: ");
c = GetInteger;
z = (b*b-4*a*c)
x = ((-b - z)/(2*a));
y = ((-b + z)/(2*a));
if (x = y)
{
printf("There is one solution: %d", x);
}
else if (x != y)
{
printf("There are two solutions: %d and %d", x, y);
}
else
{
printf("There are no solutions");
}
system("pause");
} _________________________
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Join Date: Oct 2007
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Posting Whiz in Training
There could be a possibility that if "GetInteger" is a function it is not of integer type. Otherwise if "GetInteger" is not a function then it is not declared 
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Join Date: Dec 2006
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Posting Whiz in Training
pLike, you should really consider looking at on how to indentent you code first. Well the error message you get there is basiclly specifying that GetInteger cannot be assigned to the interger. It donst know what the RHS is? Seems to me like it is function or is it macro if there was no brackets ????
And the in the if statment, there two two equals opeator not the assignment operator
ssharish
C Syntax (Toggle Plain Text)
And the in the if statment, there two two equals opeator not the assignment operator
if ( x == y )
ssharish
Last edited by ssharish2005 : Jul 24th, 2008 at 2:50 pm.
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Light Poster
Ok... Thanks that solved that, but i have a new problem, i can't get the decimals with the number here is the improved program.
#include <iostream>
#include <stdio.h>
#include "simpio.h"
#include "strlib.h"
#include "random.h"
#include "math.h"
using namespace std;
int main()
{
int a, b, c, x, y, d;
printf("Enter A: ");
a = GetInteger();
printf("Enter B: ");
b = GetInteger();
printf("Enter C: ");
c = GetInteger();
d = (b*b-4*a*c)^(1/2);
x = ((-b - d)/(2.0*a));
y = ((-b + d)/(2.0*a));
if (x == y)
printf("There is one solution: %d\n", x);
else if (x != y)
printf("There are two solutions: %d and %d\n", x, y);
else if (d <= 0)
printf("There are no solutions.\n");
else
printf("There are no solutions.\n");
system("pause");
return 0;
} _________________________
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Join Date: Jan 2008
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iostream isn't in C. I think you have quotes where you want <> in your #include for math.h, and you can't use "using namespace std" in C. Maybe you're not using a C compiler.
The problem with the decimals is that you've declared all of your variables as integers and integers can't hold decimals. Try declaring x, y, d as doubles or floats.
I'm also suspicious of this line:
^ is the bitwise XOR operator and I think (1 / 2) is going to be 0 due to integer division.
The problem with the decimals is that you've declared all of your variables as integers and integers can't hold decimals. Try declaring x, y, d as doubles or floats.
I'm also suspicious of this line:
d = (b*b-4*a*c)^(1/2);
^ is the bitwise XOR operator and I think (1 / 2) is going to be 0 due to integer division.
Last edited by VernonDozier : Jul 25th, 2008 at 12:01 am.
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Junior Poster
You're getting C++ errors. C++ is a lot stricter. Try using a C compiler. 
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Originally Posted by Clockowl 
You're getting C++ errors. C++ is a lot stricter. Try using a C compiler.
You have that backwards. These lines:
#include <iostream> using namespace std;
are fine with a C++ compiler, but cannot be used with a C compiler.
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Junior Poster
"cannot convert from" is a C++ (and in general, OOP) only error afaik.
Last edited by Clockowl : Jul 25th, 2008 at 10:16 am.
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>You're getting C++ errors. C++ is a lot stricter. Try using a C compiler.
So your advice is to ignore the warning and let the problem silently continue. Brilliant.
>"cannot convert from" is a C++ (and in general, OOP) only error afaik.
You get it in C as well if you do something stupid...just like C++.
So your advice is to ignore the warning and let the problem silently continue. Brilliant.
>"cannot convert from" is a C++ (and in general, OOP) only error afaik.
You get it in C as well if you do something stupid...just like C++.
Last edited by Narue : Jul 25th, 2008 at 10:25 am.
I'm a programmer. My attitude starts with arrogance, holds steady at condescension, and ends with hostility. Get used to it.
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Originally Posted by Clockowl
"cannot convert from" is a C++ (and in general, OOP) only error afaik.
Ah, OK, you are referring to these lines, perhaps?
x = ((-b - d)/(2.0*a)); y = ((-b + d)/(2.0*a));
I get a warning in C++ (not an error) which says "Warning: Converting from 'int' to 'double'". I don't get the "cannot convert from" error. It converts it using the C++ compiler, but throws out the warning and it's the programmer's decision about whether to pay attention to the warning or not. Maybe it's an error instead of a warning if you have certain compiler flags. In this case, it's a warning that the OP should definitely pay attention to and not ignore. You're right, though, given that this is the C forum and hence the OP should be using a C compiler, there will be no warning whatsoever on these conversions, or at least there wasn't on my C compiler.
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