creating string literal out of array of chars
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0
down vote
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I want to merge 2 string literals to one
this is my function
char* merge(const char* a, const char* b)
{
char buffer[256]; // <- danger, only storage for 256 characters.
strncpy(buffer, a, sizeof(buffer));
strncat(buffer, " ", sizeof(buffer));
strncat(buffer, b, sizeof(buffer));
strncat(buffer, "", sizeof(buffer));
char* x = buffer;
cout <<"ll "<<x<<endl;
for(int i=0;i<3;i++)
{
buffer[i]='p';
}
cout <<"ll "<<x<<endl;
return x;
}
Let's assume a="Michael" and b = John "
output (couts) is:
ll Michael John
ll ppphael John
I made some research and it is like that because x points to array allocated at stack. I want x to be pointing to string literal allocated in read-only section, so my output would be:
ll Michael John
ll Michael John
My question is how to make string literal out of array of chars buffer[256]={M,i,c,h,a,e,l, ,J,o,h,n,,0,0,0,0,0,0...,0}
(Don't ask why I'm doing such pointless job, it's for my studies and we are not allowed to use string type nor stl)
arrays string string-literals
asked Nov 11 at 14:47
pooo mn
13
|
show 1 more comment
up vote
0
down vote
favorite
I want to merge 2 string literals to one
this is my function
char* merge(const char* a, const char* b)
{
char buffer[256]; // <- danger, only storage for 256 characters.
strncpy(buffer, a, sizeof(buffer));
strncat(buffer, " ", sizeof(buffer));
strncat(buffer, b, sizeof(buffer));
strncat(buffer, "", sizeof(buffer));
char* x = buffer;
cout <<"ll "<<x<<endl;
for(int i=0;i<3;i++)
{
buffer[i]='p';
}
cout <<"ll "<<x<<endl;
return x;
}
Let's assume a="Michael" and b = John "
output (couts) is:
ll Michael John
ll ppphael John
I made some research and it is like that because x points to array allocated at stack. I want x to be pointing to string literal allocated in read-only section, so my output would be:
ll Michael John
ll Michael John
My question is how to make string literal out of array of chars buffer[256]={M,i,c,h,a,e,l, ,J,o,h,n,,0,0,0,0,0,0...,0}
(Don't ask why I'm doing such pointless job, it's for my studies and we are not allowed to use string type nor stl)
arrays string string-literals
asked Nov 11 at 14:47
pooo mn
13
String literals are part of the source code. They're not something you manipulate at runtime.
– Jonathon Reinhart
Nov 11 at 14:48
This is not C. Please don't spam tags.
– Passer By
Nov 11 at 14:49
This looks very much like C. In modern C++ this is not how this would be written.
– Jesper Juhl
Nov 11 at 14:50
1
In C++ a "literal string" is really a constant array of characters with a null-terminator. In C the arrays aren't constant, but only read-only (i.e. you still should consider them constant).
– Some programmer dude
Nov 11 at 14:50
There are also quite a few problems with the code you show, including returning pointers to local variables, and that the string inbuffercould be missing a terminator.
– Some programmer dude
Nov 11 at 14:52
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to merge 2 string literals to one
this is my function
char* merge(const char* a, const char* b)
{
char buffer[256]; // <- danger, only storage for 256 characters.
strncpy(buffer, a, sizeof(buffer));
strncat(buffer, " ", sizeof(buffer));
strncat(buffer, b, sizeof(buffer));
strncat(buffer, "", sizeof(buffer));
char* x = buffer;
cout <<"ll "<<x<<endl;
for(int i=0;i<3;i++)
{
buffer[i]='p';
}
cout <<"ll "<<x<<endl;
return x;
}
Let's assume a="Michael" and b = John "
output (couts) is:
ll Michael John
ll ppphael John
I made some research and it is like that because x points to array allocated at stack. I want x to be pointing to string literal allocated in read-only section, so my output would be:
ll Michael John
ll Michael John
My question is how to make string literal out of array of chars buffer[256]={M,i,c,h,a,e,l, ,J,o,h,n,,0,0,0,0,0,0...,0}
(Don't ask why I'm doing such pointless job, it's for my studies and we are not allowed to use string type nor stl)
arrays string string-literals
asked Nov 11 at 14:47
pooo mn
13
I want to merge 2 string literals to one
this is my function
char* merge(const char* a, const char* b)
{
char buffer[256]; // <- danger, only storage for 256 characters.
strncpy(buffer, a, sizeof(buffer));
strncat(buffer, " ", sizeof(buffer));
strncat(buffer, b, sizeof(buffer));
strncat(buffer, "", sizeof(buffer));
char* x = buffer;
cout <<"ll "<<x<<endl;
for(int i=0;i<3;i++)
{
buffer[i]='p';
}
cout <<"ll "<<x<<endl;
return x;
}
Let's assume a="Michael" and b = John "
output (couts) is:
ll Michael John
ll ppphael John
I made some research and it is like that because x points to array allocated at stack. I want x to be pointing to string literal allocated in read-only section, so my output would be:
ll Michael John
ll Michael John
My question is how to make string literal out of array of chars buffer[256]={M,i,c,h,a,e,l, ,J,o,h,n,,0,0,0,0,0,0...,0}
(Don't ask why I'm doing such pointless job, it's for my studies and we are not allowed to use string type nor stl)
arrays string string-literals
arrays string string-literals
asked Nov 11 at 14:47
pooo mn
13
asked Nov 11 at 14:47
pooo mn
13
edited Nov 11 at 14:51
asked Nov 11 at 14:47
pooo mn
13
asked Nov 11 at 14:47
pooo mn
13
asked Nov 11 at 14:47
pooo mn
13
13
String literals are part of the source code. They're not something you manipulate at runtime.
– Jonathon Reinhart
Nov 11 at 14:48
This is not C. Please don't spam tags.
– Passer By
Nov 11 at 14:49
This looks very much like C. In modern C++ this is not how this would be written.
– Jesper Juhl
Nov 11 at 14:50
1
In C++ a "literal string" is really a constant array of characters with a null-terminator. In C the arrays aren't constant, but only read-only (i.e. you still should consider them constant).
– Some programmer dude
Nov 11 at 14:50
There are also quite a few problems with the code you show, including returning pointers to local variables, and that the string inbuffercould be missing a terminator.
– Some programmer dude
Nov 11 at 14:52
|
show 1 more comment
String literals are part of the source code. They're not something you manipulate at runtime.
– Jonathon Reinhart
Nov 11 at 14:48
This is not C. Please don't spam tags.
– Passer By
Nov 11 at 14:49
This looks very much like C. In modern C++ this is not how this would be written.
– Jesper Juhl
Nov 11 at 14:50
1
In C++ a "literal string" is really a constant array of characters with a null-terminator. In C the arrays aren't constant, but only read-only (i.e. you still should consider them constant).
– Some programmer dude
Nov 11 at 14:50
There are also quite a few problems with the code you show, including returning pointers to local variables, and that the string inbuffercould be missing a terminator.
– Some programmer dude
Nov 11 at 14:52
String literals are part of the source code. They're not something you manipulate at runtime.
– Jonathon Reinhart
Nov 11 at 14:48
String literals are part of the source code. They're not something you manipulate at runtime.
– Jonathon Reinhart
Nov 11 at 14:48
This is not C. Please don't spam tags.
– Passer By
Nov 11 at 14:49
This is not C. Please don't spam tags.
– Passer By
Nov 11 at 14:49
This looks very much like C. In modern C++ this is not how this would be written.
– Jesper Juhl
Nov 11 at 14:50
This looks very much like C. In modern C++ this is not how this would be written.
– Jesper Juhl
Nov 11 at 14:50
1
1
In C++ a "literal string" is really a constant array of characters with a null-terminator. In C the arrays aren't constant, but only read-only (i.e. you still should consider them constant).
– Some programmer dude
Nov 11 at 14:50
In C++ a "literal string" is really a constant array of characters with a null-terminator. In C the arrays aren't constant, but only read-only (i.e. you still should consider them constant).
– Some programmer dude
Nov 11 at 14:50
There are also quite a few problems with the code you show, including returning pointers to local variables, and that the string in
buffer could be missing a terminator.– Some programmer dude
Nov 11 at 14:52
There are also quite a few problems with the code you show, including returning pointers to local variables, and that the string in
buffer could be missing a terminator.– Some programmer dude
Nov 11 at 14:52
|
show 1 more comment
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String literals are part of the source code. They're not something you manipulate at runtime.
– Jonathon Reinhart
Nov 11 at 14:48
This is not C. Please don't spam tags.
– Passer By
Nov 11 at 14:49
This looks very much like C. In modern C++ this is not how this would be written.
– Jesper Juhl
Nov 11 at 14:50
1
In C++ a "literal string" is really a constant array of characters with a null-terminator. In C the arrays aren't constant, but only read-only (i.e. you still should consider them constant).
– Some programmer dude
Nov 11 at 14:50
There are also quite a few problems with the code you show, including returning pointers to local variables, and that the string in
buffercould be missing a terminator.– Some programmer dude
Nov 11 at 14:52