Equals returning false in c++

up vote
7
down vote

favorite

I'm fairly new to cpp and I am trying to do a project. It says that the code must take in a filename as an argument and will be run by:

./main -i filename

I have written a for-loop that will iterate through the list of arguments to find the "-i" argument so that I can determine the filename. But this line always return false:

argv[i] == "-i"

Below is my code:

#include <string>
#include <iostream>

int main(int argc, char *argv) 
 std::string test = argv[0];
 for(int i = 0; i < argc; i++)
 if(argv[i] == "-i")
 test = argv[i+1];
 break;
 
 
 std::cout << test;
 return 1;

edited 1 hour ago

Martin Bonner

22.9k33059

asked 1 hour ago

Arjun C

434

4

Welcome to Stack Overflow! This is a very simple mistake in C++, but congratulations in writing a clear question with a Minimal, Complete, and Verifiable example.
â€“Â Martin Bonner
1 hour ago

I understand you are writing your own parser for learning purposes. Once you know how to do it and you understand the path to achieve it, have a look here. gnu.org/software/libc/manual/html_node/â€¦
â€“Â Al Bundy
14 mins ago

add a commentÂ |Â

up vote
7
down vote

favorite

I'm fairly new to cpp and I am trying to do a project. It says that the code must take in a filename as an argument and will be run by:

./main -i filename

I have written a for-loop that will iterate through the list of arguments to find the "-i" argument so that I can determine the filename. But this line always return false:

argv[i] == "-i"

Below is my code:

#include <string>
#include <iostream>

int main(int argc, char *argv) 
 std::string test = argv[0];
 for(int i = 0; i < argc; i++)
 if(argv[i] == "-i")
 test = argv[i+1];
 break;
 
 
 std::cout << test;
 return 1;

edited 1 hour ago

Martin Bonner

22.9k33059

asked 1 hour ago

Arjun C

434

4

Welcome to Stack Overflow! This is a very simple mistake in C++, but congratulations in writing a clear question with a Minimal, Complete, and Verifiable example.
â€“Â Martin Bonner
1 hour ago

I understand you are writing your own parser for learning purposes. Once you know how to do it and you understand the path to achieve it, have a look here. gnu.org/software/libc/manual/html_node/â€¦
â€“Â Al Bundy
14 mins ago

add a commentÂ |Â

up vote
7
down vote

favorite

I'm fairly new to cpp and I am trying to do a project. It says that the code must take in a filename as an argument and will be run by:

./main -i filename

I have written a for-loop that will iterate through the list of arguments to find the "-i" argument so that I can determine the filename. But this line always return false:

argv[i] == "-i"

Below is my code:

#include <string>
#include <iostream>

int main(int argc, char *argv) 
 std::string test = argv[0];
 for(int i = 0; i < argc; i++)
 if(argv[i] == "-i")
 test = argv[i+1];
 break;
 
 
 std::cout << test;
 return 1;

edited 1 hour ago

Martin Bonner

22.9k33059

asked 1 hour ago

Arjun C

434

I'm fairly new to cpp and I am trying to do a project. It says that the code must take in a filename as an argument and will be run by:

./main -i filename

I have written a for-loop that will iterate through the list of arguments to find the "-i" argument so that I can determine the filename. But this line always return false:

argv[i] == "-i"

Below is my code:

#include <string>
#include <iostream>

int main(int argc, char *argv) 
 std::string test = argv[0];
 for(int i = 0; i < argc; i++)
 if(argv[i] == "-i")
 test = argv[i+1];
 break;
 
 
 std::cout << test;
 return 1;

c++ g++ command-line-arguments

edited 1 hour ago

Martin Bonner

22.9k33059

asked 1 hour ago

Arjun C

434

edited 1 hour ago

Martin Bonner

22.9k33059

asked 1 hour ago

Arjun C

434

edited 1 hour ago

Martin Bonner

22.9k33059

edited 1 hour ago

Martin Bonner

22.9k33059

edited 1 hour ago

Martin Bonner

22.9k33059

asked 1 hour ago

Arjun C

434

asked 1 hour ago

Arjun C

434

asked 1 hour ago

Arjun C

434

4

Welcome to Stack Overflow! This is a very simple mistake in C++, but congratulations in writing a clear question with a Minimal, Complete, and Verifiable example.
â€“Â Martin Bonner
1 hour ago

I understand you are writing your own parser for learning purposes. Once you know how to do it and you understand the path to achieve it, have a look here. gnu.org/software/libc/manual/html_node/â€¦
â€“Â Al Bundy
14 mins ago

add a commentÂ |Â

4

Welcome to Stack Overflow! This is a very simple mistake in C++, but congratulations in writing a clear question with a Minimal, Complete, and Verifiable example.
â€“Â Martin Bonner
1 hour ago

I understand you are writing your own parser for learning purposes. Once you know how to do it and you understand the path to achieve it, have a look here. gnu.org/software/libc/manual/html_node/â€¦
â€“Â Al Bundy
14 mins ago

Welcome to Stack Overflow! This is a very simple mistake in C++, but congratulations in writing a clear question with a Minimal, Complete, and Verifiable example.
â€“Â Martin Bonner
1 hour ago

I understand you are writing your own parser for learning purposes. Once you know how to do it and you understand the path to achieve it, have a look here. gnu.org/software/libc/manual/html_node/â€¦
â€“Â Al Bundy
14 mins ago

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
4
down vote

accepted

argv[i] == "-i"

It compares two pointers so the result is always false.

You can fix it in multiple ways, for example:

const auto arg = std::string "-i" ;

for(int i = 0; i < argc; i++)
 if(argv[i] == arg)
 test = argv[i+1];
 break;

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

You really ought to point out the out-of-bounds error if the code is called without a filename following the -i. Probably the best way is to test against argc-1.
â€“Â Martin Bonner
1 hour ago

It'a also a good idea to start with index 1 as 0 is used for the program name.
â€“Â Thomas
18 mins ago

add a commentÂ |Â

up vote
6
down vote

You cannot compare pointers to char to string literals (char const*) using ==. Use std::strcmp() (<cstring>) or construct a std::string (<string>) from it to make it comparable to a char* using ==.

Also, given

for(int i = 0; i < argc; i++) 
 if(argv[i] == "-i") // i might be argc-1
 test = argv[i+1]; // i+1 == argc ==> boom!
 break;

if "-i" is the last element in the array argv your code will access argv out of bounds.

edited 1 hour ago

answered 1 hour ago

Swordfish

3,576727

Simplest fix to i out of range is to use i < argc-1 as the limit
â€“Â Martin Bonner
1 hour ago

or use std::string_view (C++17).
â€“Â Jarod42
1 hour ago

add a commentÂ |Â

up vote
0
down vote

try this:::

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char *argv) 
string test;
for(int i = 0; i < argc; i++) 
 cout<<"n"<<argv[i]<<endl;
 if((string)argv[i] == "-i")
 test = argv[i+1];
 cout<<"test= "<<test<<endl;
 break;
 

cout << test<<endl;
system("pause");
return 0;

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

New contributor

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
4
down vote

accepted

argv[i] == "-i"

It compares two pointers so the result is always false.

You can fix it in multiple ways, for example:

const auto arg = std::string "-i" ;

for(int i = 0; i < argc; i++)
 if(argv[i] == arg)
 test = argv[i+1];
 break;

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

You really ought to point out the out-of-bounds error if the code is called without a filename following the -i. Probably the best way is to test against argc-1.
â€“Â Martin Bonner
1 hour ago

It'a also a good idea to start with index 1 as 0 is used for the program name.
â€“Â Thomas
18 mins ago

add a commentÂ |Â

up vote
4
down vote

accepted

argv[i] == "-i"

It compares two pointers so the result is always false.

You can fix it in multiple ways, for example:

const auto arg = std::string "-i" ;

for(int i = 0; i < argc; i++)
 if(argv[i] == arg)
 test = argv[i+1];
 break;

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

You really ought to point out the out-of-bounds error if the code is called without a filename following the -i. Probably the best way is to test against argc-1.
â€“Â Martin Bonner
1 hour ago

It'a also a good idea to start with index 1 as 0 is used for the program name.
â€“Â Thomas
18 mins ago

add a commentÂ |Â

up vote
4
down vote

accepted

argv[i] == "-i"

It compares two pointers so the result is always false.

You can fix it in multiple ways, for example:

const auto arg = std::string "-i" ;

for(int i = 0; i < argc; i++)
 if(argv[i] == arg)
 test = argv[i+1];
 break;

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

argv[i] == "-i"

It compares two pointers so the result is always false.

You can fix it in multiple ways, for example:

const auto arg = std::string "-i" ;

for(int i = 0; i < argc; i++)
 if(argv[i] == arg)
 test = argv[i+1];
 break;

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

answered 1 hour ago

Edgar RokjÃ„Ân

14.5k42650

You really ought to point out the out-of-bounds error if the code is called without a filename following the -i. Probably the best way is to test against argc-1.
â€“Â Martin Bonner
1 hour ago

It'a also a good idea to start with index 1 as 0 is used for the program name.
â€“Â Thomas
18 mins ago

add a commentÂ |Â

You really ought to point out the out-of-bounds error if the code is called without a filename following the -i. Probably the best way is to test against argc-1.
â€“Â Martin Bonner
1 hour ago

It'a also a good idea to start with index 1 as 0 is used for the program name.
â€“Â Thomas
18 mins ago

You really ought to point out the out-of-bounds error if the code is called without a filename following the -i. Probably the best way is to test against argc-1.
â€“Â Martin Bonner
1 hour ago

It'a also a good idea to start with index 1 as 0 is used for the program name.
â€“Â Thomas
18 mins ago

add a commentÂ |Â

up vote
6
down vote

Also, given

for(int i = 0; i < argc; i++) 
 if(argv[i] == "-i") // i might be argc-1
 test = argv[i+1]; // i+1 == argc ==> boom!
 break;

if "-i" is the last element in the array argv your code will access argv out of bounds.

edited 1 hour ago

answered 1 hour ago

Swordfish

3,576727

Simplest fix to i out of range is to use i < argc-1 as the limit
â€“Â Martin Bonner
1 hour ago

or use std::string_view (C++17).
â€“Â Jarod42
1 hour ago

add a commentÂ |Â

up vote
6
down vote

Also, given

for(int i = 0; i < argc; i++) 
 if(argv[i] == "-i") // i might be argc-1
 test = argv[i+1]; // i+1 == argc ==> boom!
 break;

if "-i" is the last element in the array argv your code will access argv out of bounds.

edited 1 hour ago

answered 1 hour ago

Swordfish

3,576727

Simplest fix to i out of range is to use i < argc-1 as the limit
â€“Â Martin Bonner
1 hour ago

or use std::string_view (C++17).
â€“Â Jarod42
1 hour ago

add a commentÂ |Â

up vote
6
down vote

Also, given

for(int i = 0; i < argc; i++) 
 if(argv[i] == "-i") // i might be argc-1
 test = argv[i+1]; // i+1 == argc ==> boom!
 break;

if "-i" is the last element in the array argv your code will access argv out of bounds.

edited 1 hour ago

answered 1 hour ago

Swordfish

3,576727

Also, given

for(int i = 0; i < argc; i++) 
 if(argv[i] == "-i") // i might be argc-1
 test = argv[i+1]; // i+1 == argc ==> boom!
 break;

if "-i" is the last element in the array argv your code will access argv out of bounds.

edited 1 hour ago

answered 1 hour ago

Swordfish

3,576727

edited 1 hour ago

answered 1 hour ago

Swordfish

3,576727

answered 1 hour ago

Swordfish

3,576727

answered 1 hour ago

Swordfish

3,576727

Simplest fix to i out of range is to use i < argc-1 as the limit
â€“Â Martin Bonner
1 hour ago

or use std::string_view (C++17).
â€“Â Jarod42
1 hour ago

add a commentÂ |Â

Simplest fix to i out of range is to use i < argc-1 as the limit
â€“Â Martin Bonner
1 hour ago

or use std::string_view (C++17).
â€“Â Jarod42
1 hour ago

Simplest fix to i out of range is to use i < argc-1 as the limit
â€“Â Martin Bonner
1 hour ago

or use std::string_view (C++17).
â€“Â Jarod42
1 hour ago

add a commentÂ |Â

up vote
0
down vote

try this:::

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char *argv) 
string test;
for(int i = 0; i < argc; i++) 
 cout<<"n"<<argv[i]<<endl;
 if((string)argv[i] == "-i")
 test = argv[i+1];
 cout<<"test= "<<test<<endl;
 break;
 

cout << test<<endl;
system("pause");
return 0;

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

New contributor

add a commentÂ |Â

up vote
0
down vote

try this:::

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char *argv) 
string test;
for(int i = 0; i < argc; i++) 
 cout<<"n"<<argv[i]<<endl;
 if((string)argv[i] == "-i")
 test = argv[i+1];
 cout<<"test= "<<test<<endl;
 break;
 

cout << test<<endl;
system("pause");
return 0;

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

New contributor

add a commentÂ |Â

up vote
0
down vote

try this:::

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char *argv) 
string test;
for(int i = 0; i < argc; i++) 
 cout<<"n"<<argv[i]<<endl;
 if((string)argv[i] == "-i")
 test = argv[i+1];
 cout<<"test= "<<test<<endl;
 break;
 

cout << test<<endl;
system("pause");
return 0;

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

New contributor

try this:::

#include <iostream>
#include <string>
using namespace std;
int main(int argc, char *argv) 
string test;
for(int i = 0; i < argc; i++) 
 cout<<"n"<<argv[i]<<endl;
 if((string)argv[i] == "-i")
 test = argv[i+1];
 cout<<"test= "<<test<<endl;
 break;
 

cout << test<<endl;
system("pause");
return 0;

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

New contributor

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

New contributor

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

answered 40 mins ago

ElbÃ¦dry JÃ¸ba

New contributor

ElbÃ¦dry JÃ¸ba is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a commentÂ |Â

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