•
•
•
•
What is DaniWeb IT Discussion Community?
You're currently browsing the C section within the Software Development category of DaniWeb, a massive community of 391,769 software developers, web developers, Internet marketers, and tech gurus who are all enthusiastic about making contacts, networking, and learning from each other. In fact, there are 3,185 IT professionals currently interacting right now! Registration is free, only takes a minute and lets you enjoy all of the interactive features of the site.
Please support our C advertiser:
Views: 8500 | Replies: 9
 |
•
•
Join Date: Aug 2004
Posts: 3
Reputation: 
Rep Power: 0
Solved Threads: 0
Newbie Poster
cannot convert from 'double' to 'float [6][3]' Using this code:
Fees = ovrh + (wr * z);
I am getting the following error:
error C2440: '=' : cannot convert from 'double' to 'float [6][3]'
Fees, ovrhd, wr and z are all declared as float. I believe if I can figure out what the problem is, I will have a functioning program. After several days straight of this program, I believe that I may be forgetting something basic.
Fees = ovrh + (wr * z);
I am getting the following error:
error C2440: '=' : cannot convert from 'double' to 'float [6][3]'
Fees, ovrhd, wr and z are all declared as float. I believe if I can figure out what the problem is, I will have a functioning program. After several days straight of this program, I believe that I may be forgetting something basic.
•
•
Join Date: Apr 2004
Posts: 3,449
Reputation: 
Rep Power: 16
Solved Threads: 138
Could you show more of the code? Specifically the actual declarations for the variables would be helpful. I find no such issue with this minimal code sample.
#include <stdio.h>
int main(void)
{
float ovrh = 1.0F, wr = 1.0F, z = 1.0F, Fees = ovrh + (wr * z);
printf("Fees = %g\n", Fees);
return 0;
} •
•
Join Date: Jun 2004
Location: Marin, CA, USA
Posts: 434
Reputation:
Rep Power: 5
Solved Threads: 10
Unprevaricator
The calculation of two floats yields a double. Compilers often WARN about converting from double to float because of a possible loss of precision.
YOUR error code sure sounds like you have an ARRAY of floats like
float fees[6][3];
fees = 1.0 * 2.0; // error double to float [6][3]
YOUR error code sure sounds like you have an ARRAY of floats like
float fees[6][3];
fees = 1.0 * 2.0; // error double to float [6][3]
•
•
Join Date: Apr 2004
Posts: 3,449
Reputation:
Rep Power: 16
Solved Threads: 138
Slight nitpick:
Those aren't two floats (1.0 and 2.0), they're two doubles.
•
•
•
•
Originally Posted by Chainsaw
The calculation of two floats yields a double.
fees = 1.0 * 2.0;
•
•
Join Date: Aug 2004
Posts: 3
Reputation:
Rep Power: 0
Solved Threads: 0
Newbie Poster
The purpose of the program is to store shipping costs in an array and then display them on the screen, in a formatted output.
Here are the declarations:
float Fees [6][3];
float i, j;
int wr, z;
const double ovrh = 2.27; // I tried to enter this as a const float, but the compiler didn't like it, so I changed it to a double which solved that one problem.
The assignment is 4 days late, so I am going to just turn it in as it. I would still like to see it work though. Thank you so much for your help!
Here are the declarations:
float Fees [6][3];
float i, j;
int wr, z;
const double ovrh = 2.27; // I tried to enter this as a const float, but the compiler didn't like it, so I changed it to a double which solved that one problem.
The assignment is 4 days late, so I am going to just turn it in as it. I would still like to see it work though. Thank you so much for your help!
•
•
•
•
Originally Posted by Dave Sinkula
Could you show more of the code? Specifically the actual declarations for the variables would be helpful. I find no such issue with this minimal code sample.
#include <stdio.h> int main(void) { float ovrh = 1.0F, wr = 1.0F, z = 1.0F, Fees = ovrh + (wr * z); printf("Fees = %g\n", Fees); return 0; }
•
•
Join Date: Aug 2004
Posts: 3
Reputation:
Rep Power: 0
Solved Threads: 0
Newbie Poster
The purpose of the program is to store shipping costs in an array and then display them on the screen, in a formatted output.
Here are the declarations:
float Fees [6][3];
float i, j;
int wr, z;
const double ovrh = 2.27; // I tried to enter this as a const float, but the compiler didn't like it, so I changed it to a double which solved that one problem.
The assignment is 4 days late, so I am going to just turn it in as it. Thank you so much for your help!
Here are the declarations:
float Fees [6][3];
float i, j;
int wr, z;
const double ovrh = 2.27; // I tried to enter this as a const float, but the compiler didn't like it, so I changed it to a double which solved that one problem.
The assignment is 4 days late, so I am going to just turn it in as it. Thank you so much for your help!
•
•
•
•
Originally Posted by Chainsaw
The calculation of two floats yields a double. Compilers often WARN about converting from double to float because of a possible loss of precision.
YOUR error code sure sounds like you have an ARRAY of floats like
float fees[6][3];
fees = 1.0 * 2.0; // error double to float [6][3]
•
•
Join Date: Apr 2004
Posts: 3,449
Reputation:
Rep Power: 16
Solved Threads: 138
Aha! So your original statement was not entirely accurate. Fees is declared as an array of an array of float -- this is not the same and your compiler was kindly informing you of this. A single float might be Fees[0][0]. Try this (I'm assuming i and j are loop indices -- except that then they must be integral types and not floating point, but I'm used to i and j as loop indices -- and this is inside these loops.
•
•
•
•
Fees, ovrhd, wr and z are all declared as float.
Fees[i][j] = ovrh + (wr * z);
•
•
Join Date: Jun 2004
Location: Marin, CA, USA
Posts: 434
Reputation:
Rep Power: 5
Solved Threads: 10
Unprevaricator
Ah HAH! As I suspected.
Dave, you are right that '1.0' is a double. However, I still believe that two floats used in a calculation result in a double, in the same way that two chars used in a calculation result in an int. In any case, the compiler can deal with that and generally puts out a warning, not an error.
Dave, you are right that '1.0' is a double. However, I still believe that two floats used in a calculation result in a double, in the same way that two chars used in a calculation result in an int. In any case, the compiler can deal with that and generally puts out a warning, not an error.
•
•
Join Date: Apr 2004
Posts: 3,449
Reputation:
Rep Power: 16
Solved Threads: 138
•
•
•
•
Originally Posted by Chainsaw
Ah HAH! As I suspected.
Dave, you are right that '1.0' is a double. However, I still believe that two floats used in a calculation result in a double, in the same way that two chars used in a calculation result in an int. In any case, the compiler can deal with that and generally puts out a warning, not an error.
const double ovrh = 2.27;
•
•
Join Date: Apr 2004
Posts: 3,449
Reputation:
Rep Power: 16
Solved Threads: 138
Oops. I didn't read the comment. It makes me curious -- tnorton: are you saying that the compiler barked about something like this? But the other stuff about the promotions should be correct.
•
•
•
•
// I tried to enter this as a const float, but the compiler didn't like it, so I changed it to a double which solved that one problem
const float ovrh = 2.27F;
|
•
•
•
•
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)
•
•
•
•
•
•
•
•
DaniWeb C Marketplace
Linear Mode
Similar Threads
Other Threads in the C Forum
- Previous Thread: Commission or Brokerage Problem
- Next Thread: Linking errors of socket functions
