hi
another simple problem sorry. i've got a string, "Buffer", and an int, Loop.
Somehow, this:
Buffer.length<=Loop+1
gives this:
error C2296: '<=' : illegal, left operand has type 'unsigned int (__thiscall
std::basic_string<char,struct std::char_traits<char>,class std::allocato
r<char> >::*'
Buffer.length should return an int, correct? loop is certainly an int. what
am i doing wrong?
cheers
dave
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| User: "Wouter Lievens" |
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| Title: Re: type prob |
27 Nov 2003 02:12:15 AM |
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"Dave" <davefromalbury_REMOVE_THIS_TO_EMAIL_@hotmail.com> schreef in bericht
news:3fc57821@dnews.tpgi.com.au...
hi
another simple problem sorry. i've got a string, "Buffer", and an int,
Loop.
Somehow, this:
Buffer.length<=Loop+1
gives this:
error C2296: '<=' : illegal, left operand has type 'unsigned int
(__thiscall
std::basic_string<char,struct std::char_traits<char>,class std::allocato
r<char> >::*'
Buffer.length should return an int, correct? loop is certainly an int.
what
am i doing wrong?
cheers
dave
Though compile errors seem cryptic, they mostly speak the truth.
Left operand in this case has a function's type: that's becasue you're not
calling it!
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| User: "Dave" |
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| Title: Re: type prob |
28 Nov 2003 12:01:29 AM |
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sorry i think i'm missing something everone else is seeing. how do i find
the length of the string "Buffer", as an "int" data type?
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| User: "Rolf Magnus" |
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| Title: Re: type prob |
27 Nov 2003 05:22:57 AM |
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Dave wrote:
sorry i think i'm missing something everone else is seeing. how do i
find the length of the string "Buffer", as an "int" data type?
You got the answer already. You must _call_ the function to get its
result. You're comparing the function itself to Loop+1, not the result
of a call to it. Buffer.length gives you the function. Buffer.length()
calls the function and gives you the result.
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| User: "Victor Bazarov" |
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| Title: Re: type prob |
27 Nov 2003 11:31:45 AM |
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"Rolf Magnus" <ramagnus@t-online.de> wrote...
Dave wrote:
sorry i think i'm missing something everone else is seeing. how do i
find the length of the string "Buffer", as an "int" data type?
You got the answer already. You must _call_ the function to get its
result. You're comparing the function itself to Loop+1, not the result
of a call to it. Buffer.length gives you the function. Buffer.length()
calls the function and gives you the result.
Thanks, Rolf.
Now, for the inattentive ones: it's the parentheses.
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| User: "Dave" |
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| Title: Re: type prob |
28 Nov 2003 09:15:54 AM |
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Ahahaha yes i am feeling stupid. Looks like its back to basics for me:
10 PRINT "HELLO WORLD"
20 GOTO 10
cheers
dave
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| User: "Josephine Schafer" |
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| Title: Re: type prob |
26 Nov 2003 10:34:39 PM |
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"Dave" <davefromalbury_REMOVE_THIS_TO_EMAIL_@hotmail.com> wrote in message
news:3fc57821@dnews.tpgi.com.au...
hi
another simple problem sorry. i've got a string, "Buffer", and an int, Loop.
Somehow, this:
Buffer.length<=Loop+1
gives this:
error C2296: '<=' : illegal, left operand has type 'unsigned int (__thiscall
std::basic_string<char,struct std::char_traits<char>,class std::allocato
r<char> >::*'
Buffer.length should return an int, correct? loop is certainly an int. what
am i doing wrong?
No.
Buffer.length returns std::string::size_type and not int.
Try this -
#include <string>
int main()
{
std::string Buffer("abc");
std::string::size_type loop = Buffer.length ();
}
HTH,
J.Schafer
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| User: "Victor Bazarov" |
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| Title: Re: type prob |
26 Nov 2003 10:35:55 PM |
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"Dave" <davefromalbury_REMOVE_THIS_TO_EMAIL_@hotmail.com> wrote...
another simple problem sorry. i've got a string, "Buffer", and an int,
Loop.
Somehow, this:
Buffer.length<=Loop+1
Did you mean to say
Buffer.length()<=Loop+1
gives this:
error C2296: '<=' : illegal, left operand has type 'unsigned int
(__thiscall
std::basic_string<char,struct std::char_traits<char>,class std::allocato
r<char> >::*'
Buffer.length should return an int, correct? loop is certainly an int.
what
am i doing wrong?
You're not calling the function, you're placing its name in the
expression. That's what the compiler is telling you: "Left operand
has type '...<function declaration>...' "
Victor
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