maybe its time to learn and stop being a useless Black person.
You are actually the useless one.
If someone asks a question, you don't tell them to go somewhere else your mentally braindead moron. Learning is exactly what I am doing right now you homosexual, this is impossible to learn without outside guidance, this is a moron system built by morons like you, so either explain it or shut the frick up you worthless sack of shit.
2 years ago
Anonymous
>learning is what I am doing right now >NO I REFUSE TO USE THE TOOL DESIGNED EXACTLY TO TELL ME HOW TO SEE WHAT IS HAPPENING
google "how to use <MY IDE> <MY LANGUAGE> <DEBUGGER>"
maybe its time to learn and stop being a useless Black person.
[...]
You are actually the useless one.
If someone asks a question, you don't tell them to go somewhere else your mentally braindead moron. Learning is exactly what I am doing right now you homosexual, this is impossible to learn without outside guidance, this is a moron system built by morons like you, so either explain it or shut the frick up you worthless sack of shit.
>learning is what I am doing right now >NO I REFUSE TO USE THE TOOL DESIGNED EXACTLY TO TELL ME HOW TO SEE WHAT IS HAPPENING
google "how to use <MY IDE> <MY LANGUAGE> <DEBUGGER>"
Somebody should ban this homosexual everytime he posts, you are such a b***h.
Among other things, you calculate i3 as -4 and you return an array allocated on stack (s2) which is a clear mistake. It doesn't output s2:p for me, it outputs s2:h even with your broken code. I got it to work fully after fixing the two errors I mentioned above.
After line 35, values of indices are:
i1 = 6
i2 = 1
i3 = -4
i4 = 0
Your logic is wrong. i3 shouldn't be negative.
In the while loop on line 37:
When you copy i1 to i2, you forget to copy the ' ' character.
On line 9:
What the frick is return type *? It should be char *
And so on...
Take it slow. Do it step by step. You've gone wayyyyy ahead of yourself with this program.
Learn to step through your code by hand. Using a debugger comes after that.
Now frick off and study.
>Omg
*vomit*
2 years ago
Anonymous
correction:
when you copy s1 to s2
main.c:2:1: warning: return type defaults to ‘int’ [-Wimplicit-int]
2 | main()
| ^~~~
main.c: In function ‘main’:
main.c:5:19: warning: implicit declaration of function ‘trim’ [-Wimplicit-function-declaration]
5 | printf("%sn",trim(string));
| ^~~~
main.c:5:14: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]
5 | printf("%sn",trim(string));
| ~^ ~~~~~~~~~~~~
| | |
| | int
| char *
| %d
main.c: At top level:
main.c:9:3: warning: return type defaults to ‘int’ [-Wimplicit-int]
9 | * trim(char s1[])
| ^~~~
main.c:9:3: error: conflicting types for ‘trim’
main.c:5:19: note: previous implicit declaration of ‘trim’ was here
5 | printf("%sn",trim(string));
| ^~~~
main.c: In function ‘trim’:
main.c:44:12: warning: returning ‘char *’ from a function with incompatible return type ‘int *’ [-Wincompatible-pointer-types]
44 | return s2;
| ^~
main.c:44:12: warning: function returns address of local variable [-Wreturn-local-addr]
Yes exactly. Don't know which compiler OP is using. trim function returns a pointer to s2. And s2 doesn't have a ' ' terminating character.
OP is a confirmed moron and possibly also a troony.
2 years ago
Anonymous
contd...
OP is an actual moron.
Now that I look at this program again...
What the frick is that on line 36?
char s2[i3];???????!!!!!!!
Either dynamically declare a new array based on the length or declare the size a large constant.
And don't forget to terminate your string with a ' '.
What an absolute fricking moron.
2 years ago
Anonymous
You can do variable size length arrays in C and this is C. The mistake is returning s2, which should not be done regardless if the length is variable or constant.
2 years ago
Anonymous
contd...
OP is an actual moron.
Now that I look at this program again...
What the frick is that on line 36?
char s2[i3];???????!!!!!!!
Either dynamically declare a new array based on the length or declare the size a large constant.
And don't forget to terminate your string with a ' '.
What an absolute fricking moron.
People like you should be gunned down. You probably learnt everything from somebody else or copy pasted everything you wienersucking boomer b***h homosexual have a nice day.
2 years ago
Anonymous
contd...
OP is an actual moron.
Now that I look at this program again...
What the frick is that on line 36?
char s2[i3];???????!!!!!!!
Either dynamically declare a new array based on the length or declare the size a large constant.
And don't forget to terminate your string with a ' '.
What an absolute fricking moron.
As to why it might be printing p, one possibility is with your i3 calculated as -1, s2[i4] = s1[i2]; line could actually be modifying s2, not the data inside that array, but the actual pointer in memory. So after that assignment that points to some random place in memory, and contents at that address happen to be p. After that there is no modifying of the pointer, so all assignments work as they should.
main.c:2:1: warning: return type defaults to ‘int’ [-Wimplicit-int]
2 | main()
| ^~~~
main.c: In function ‘main’:
main.c:5:19: warning: implicit declaration of function ‘trim’ [-Wimplicit-function-declaration]
5 | printf("%sn",trim(string));
| ^~~~
main.c:5:14: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]
5 | printf("%sn",trim(string));
| ~^ ~~~~~~~~~~~~
| | |
| | int
| char *
| %d
main.c: At top level:
main.c:9:3: warning: return type defaults to ‘int’ [-Wimplicit-int]
9 | * trim(char s1[])
| ^~~~
main.c:9:3: error: conflicting types for ‘trim’
main.c:5:19: note: previous implicit declaration of ‘trim’ was here
5 | printf("%sn",trim(string));
| ^~~~
main.c: In function ‘trim’:
main.c:44:12: warning: returning ‘char *’ from a function with incompatible return type ‘int *’ [-Wincompatible-pointer-types]
44 | return s2;
| ^~
main.c:44:12: warning: function returns address of local variable [-Wreturn-local-addr]
Nigeria
Explain.
why don't you step through the dubugger you moron and see why
Frick you b***h, I don't know how to use the debugger, explain it.
No. Learn to use the debugger.
You are actually the useless one.
If someone asks a question, you don't tell them to go somewhere else your mentally braindead moron. Learning is exactly what I am doing right now you homosexual, this is impossible to learn without outside guidance, this is a moron system built by morons like you, so either explain it or shut the frick up you worthless sack of shit.
>learning is what I am doing right now
>NO I REFUSE TO USE THE TOOL DESIGNED EXACTLY TO TELL ME HOW TO SEE WHAT IS HAPPENING
google "how to use <MY IDE> <MY LANGUAGE> <DEBUGGER>"
maybe its time to learn and stop being a useless Black person.
Somebody should ban this homosexual everytime he posts, you are such a b***h.
char s2[i3];
kys
Nothing wrong with that you fricking idiot, you don't know shit so shut the frick up.
post the code as text if you want help
#include <stdio.h>
main()
{
char string[] = {' ','h','e','l','l','o',' ',' ',' '};
printf("%sn",trim(string));
}
* trim(char s1[])
{
int i1 = 0; //char array index
int i2 = 0; //start index
int i3 = 0; //word indicator
int i4 = 0; //end cut measure
while(s1[i1] != ' ')
{
if(s1[i1] == ' ' || s1[i1] == 't' || s1[i1] == 'n')
{
i3 = 0;
i4++;
}
else
{
if(i2 == 0)
{
i2 = i1;
}
i3 = 1;
i4 = 0;
}
i1++;
}
i1 = i1 - i4; //end
i3 = i2 - i1 + 1; //length
i4 = 0; //newIndex
char s2[i3];
while(i2 < i1)
{
s2[i4] = s1[i2];
printf("s2:%c, s1:%cn",s2[i4],s1[i2]);
i2++;
i4++;
}
return s2;
}
Among other things, you calculate i3 as -4 and you return an array allocated on stack (s2) which is a clear mistake. It doesn't output s2:p for me, it outputs s2:h even with your broken code. I got it to work fully after fixing the two errors I mentioned above.
Omg thank you.
After line 35, values of indices are:
i1 = 6
i2 = 1
i3 = -4
i4 = 0
Your logic is wrong. i3 shouldn't be negative.
In the while loop on line 37:
When you copy i1 to i2, you forget to copy the ' ' character.
On line 9:
What the frick is return type *? It should be char *
And so on...
Take it slow. Do it step by step. You've gone wayyyyy ahead of yourself with this program.
Learn to step through your code by hand. Using a debugger comes after that.
Now frick off and study.
>Omg
*vomit*
correction:
when you copy s1 to s2
Yes exactly. Don't know which compiler OP is using. trim function returns a pointer to s2. And s2 doesn't have a ' ' terminating character.
OP is a confirmed moron and possibly also a troony.
contd...
OP is an actual moron.
Now that I look at this program again...
What the frick is that on line 36?
char s2[i3];???????!!!!!!!
Either dynamically declare a new array based on the length or declare the size a large constant.
And don't forget to terminate your string with a ' '.
What an absolute fricking moron.
You can do variable size length arrays in C and this is C. The mistake is returning s2, which should not be done regardless if the length is variable or constant.
People like you should be gunned down. You probably learnt everything from somebody else or copy pasted everything you wienersucking boomer b***h homosexual have a nice day.
Lose some weight you fat frick.
As to why it might be printing p, one possibility is with your i3 calculated as -1, s2[i4] = s1[i2]; line could actually be modifying s2, not the data inside that array, but the actual pointer in memory. So after that assignment that points to some random place in memory, and contents at that address happen to be p. After that there is no modifying of the pointer, so all assignments work as they should.
Yep needs to be
#include <stdio.h>
main()
{
char string[] = {' ','h','e','l','l','o',' ',' ',' '};
printf("%sn",trim(string));
}
* trim(char s1[])
{
int i1 = 0; //char array index
int i2 = 0; //start index
int i3 = 0; //word indicator
int i4 = 0; //end cut measure
while(s1[i1] != ' ')
{
if(s1[i1] == ' ' || s1[i1] == 't' || s1[i1] == 'n')
{
i3 = 0;
i4++;
}
else
{
if(i2 == 0)
{
i2 = i1;
}
i3 = 1;
i4 = 0;
}
i1++;
}
i1 = i1 - i4; //end
i3 = i1 - i2 + 1; //length
i4 = 0; //newIndex
char s2[i3];
while(i2 < i1)
{
s2[i4] = s1[i2];
printf("s2:%c, s1:%cn",s2[i4],s1[i2]);
i2++;
i4++;
}
return s2;
}
cool memory management, ahmed
>not wasting time on this crap
learn to debug homosexual
Cleanup your code and maybe someone will be able to read it.
People like OP is the reason why Rust exists.
I'm pretty sure OP isn't a troony.
>I'm pretty sure OP isn't a troony.
Nice try, OP.
main.c:2:1: warning: return type defaults to ‘int’ [-Wimplicit-int]
2 | main()
| ^~~~
main.c: In function ‘main’:
main.c:5:19: warning: implicit declaration of function ‘trim’ [-Wimplicit-function-declaration]
5 | printf("%sn",trim(string));
| ^~~~
main.c:5:14: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]
5 | printf("%sn",trim(string));
| ~^ ~~~~~~~~~~~~
| | |
| | int
| char *
| %d
main.c: At top level:
main.c:9:3: warning: return type defaults to ‘int’ [-Wimplicit-int]
9 | * trim(char s1[])
| ^~~~
main.c:9:3: error: conflicting types for ‘trim’
main.c:5:19: note: previous implicit declaration of ‘trim’ was here
5 | printf("%sn",trim(string));
| ^~~~
main.c: In function ‘trim’:
main.c:44:12: warning: returning ‘char *’ from a function with incompatible return type ‘int *’ [-Wincompatible-pointer-types]
44 | return s2;
| ^~
main.c:44:12: warning: function returns address of local variable [-Wreturn-local-addr]