IDE Dev-C++

bongo264 · 11-08-2009, 12:04 AM

I am using the dev-c++ compiler and going through a C programing cd bu Mark Virtue that is quite good.

The problem I am having is I am at a point where I am checking the value of a variable and on the CD he is using a single quote around it as the value of the variable is a single character.

Now I am using the same compiler as he is (Dev-C++) but it is a different version. His version will highlight single quotes and everything in between it but mine only does this with double quotes. Also when I do compile with single quote I get the following error on the line with the single quote around a character.

[Warning] comparison between pointer and integer

any ideas???

I have tried to use double quotes but it does not work either.

nadroj · 11-08-2009, 12:34 AM

Its probably easier if you just post the small piece of code, than explain it in English. Are you doing something like this?

Code:

    char* str="a";

    if ( str == 'a')

Which will result in the warning your seeing. If not, again, post your code.

11-08-2009, 12:04 AM	#1
bongo264 Registered User Join Date: Nov 2009 Posts: 1	IDE Dev-C++ I am using the dev-c++ compiler and going through a C programing cd bu Mark Virtue that is quite good. The problem I am having is I am at a point where I am checking the value of a variable and on the CD he is using a single quote around it as the value of the variable is a single character. Now I am using the same compiler as he is (Dev-C++) but it is a different version. His version will highlight single quotes and everything in between it but mine only does this with double quotes. Also when I do compile with single quote I get the following error on the line with the single quote around a character. [Warning] comparison between pointer and integer any ideas??? I have tried to use double quotes but it does not work either.
bongo264 is offline

11-08-2009, 12:34 AM	#2
nadroj Registered User Join Date: Oct 2006 Location: Canada Posts: 848	Its probably easier if you just post the small piece of code, than explain it in English. Are you doing something like this? Code: char* str="a"; if ( str == 'a') Which will result in the warning your seeing. If not, again, post your code.
nadroj is offline

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