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merrittr
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 Posted: Mon Nov 13, 2006 10:10 am Post subject: trying to concat a filename (newbie) |
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I was trying to build a list of input files using users input
ie #>program 1 2 3 4 5 6 7 8 9 10 11 12
in the program I wanted name generated input filenames
char beginning[64]="count",endding[64]=".txt", outf[64]="data";
char cf1[64]=strcat(strcat(beginning,argv[0]),endding);
//sprintf (cf1, "%s%s%s",beginning,argv[0],endding);
ifstream in1 (cf1);
what I get at gcc time:
ptcllocate.cpp:16: error: invalid initializer
as you can see I tried sprintf too (after looking in this group) any
hints
on how I can get this to work? |
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benben
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| Posted: Mon Nov 13, 2006 10:10 am Post subject: Re: trying to concat a filename (newbie) |
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merrittr wrote:
| Quote: | I was trying to build a list of input files using users input
ie #>program 1 2 3 4 5 6 7 8 9 10 11 12
in the program I wanted name generated input filenames
char beginning[64]="count",endding[64]=".txt", outf[64]="data";
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Should have been:
const char[] beginning = "count";
const char[] endding = ".txt";
const char[] outf = "data";
| Quote: | char cf1[64]=strcat(strcat(beginning,argv[0]),endding);
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strcat returns a pointer (which points to the first element of
beginning, in this case.) However, you cannot initiate a char[64] array
with a pointer. It just doesn't make sense.
A clearer way to do it is to be explicit about the buffer:
char buff[128]; // fixed size buffer can overflow
strcat(buff, beginning);
strcat(buff, argv[0]);
strcat(buff, endding);
Google on strcat if you are not sure how it should be used.
| Quote: | //sprintf (cf1, "%s%s%s",beginning,argv[0],endding);
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There are much better alternatives, such as using the standard library
string:
std::string beginning = "count";
std::string endding = ".txt";
std::string str = beginning + argv[0] + endding;
or a string stream:
std::ostringstream oss;
oss << "count" << argv[0] << ".txt";
std::string str = oss.str();
Just avoid writing C in C++.
Regards,
Ben |
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Salt_Peter
Guest
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| Posted: Mon Nov 13, 2006 10:10 am Post subject: Re: trying to concat a filename (newbie) |
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merrittr wrote:
| Quote: | I was trying to build a list of input files using users input
ie #>program 1 2 3 4 5 6 7 8 9 10 11 12
in the program I wanted name generated input filenames
char beginning[64]="count",endding[64]=".txt", outf[64]="data";
char cf1[64]=strcat(strcat(beginning,argv[0]),endding);
//sprintf (cf1, "%s%s%s",beginning,argv[0],endding);
ifstream in1 (cf1);
what I get at gcc time:
ptcllocate.cpp:16: error: invalid initializer
as you can see I tried sprintf too (after looking in this group) any
hints
on how I can get this to work?
|
Why not use a std::vector of std::strings instead.
And usually, argv[0] is the program's path + executable name.
So you need to collect all argv[]s except the first and the program
should work for any number of arguements.
#include <iostream>
#include <string>
#include <vector>
#include <iterator>
template< typename T >
void show_parameters(std::vector< T >& r_v)
{
std::copy( r_v.begin(),
r_v.end(),
std::ostream_iterator< T >(std::cout, "\n") );
}
int main(int argc, char* argv[])
{
const size_t Size = static_cast<size_t>(argc); // cvt to size_t
std::vector< std::string > vargs(Size - 1); // container, presized
const std::string s_count("count"), s_extension(".txt"); // const s
for( size_t i = 0; i < vargs.size(); ++i)
{
vargs[i] = s_count + argv[i + 1] + s_extension;
}
show_parameters< std::string >( vargs );
}
/* program 1 2 3 4 5
count1.txt
count2.txt
count3.txt
count4.txt
count5.txt
*/ |
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