After modifying string (char*) in a function the value won't show unless printed inside function
#include <stdio.h>
#include <string.h>
void increase(char **tab)
{
int i=strlen(*tab);
int o=i;
while((*tab)[i]!='5' && i>=0)
{
i--;
}
if(i>=0)
{
int tt;
char array[o+1];
for(tt=0;tt<o;tt++)
{
if(tt!=i)
array[tt]=(*tab)[tt];
else
array[tt]='6';
}
array[o]='';
*tab=array;
// printf("n%s",*tab);
}
else
{
char array[o+2];
int tt;
for(tt=0;tt<=o;tt++)
{
array[tt]='5';
}
array[o+1]='';
*tab=array;
// printf("n%s",*tab);
}
}
int main()
{
int n;
char *test;
test="555";
increase(&test);
printf("n%s",test);
return 0;
}
Okay, so increase() is meant to replace number in test with the next number containing only 5 and 6.
What I wanted to do is to change the value of test directly by using pointer to char*. Everything seems to work just fine untill it comes to displaying changed value - it simply won't show unless was asked to do it inside the
increase() function. Once I add
printf("n%s",*tab);
inside any of conditions (commented), everything works just fine (excluding showing a double result).
What causes a problem here?
555 is just a testing value, actually any number made of 5s or 6s will do the work.
c arrays string pointers char
asked Nov 12 at 1:20
Igor
193
add a comment |
#include <stdio.h>
#include <string.h>
void increase(char **tab)
{
int i=strlen(*tab);
int o=i;
while((*tab)[i]!='5' && i>=0)
{
i--;
}
if(i>=0)
{
int tt;
char array[o+1];
for(tt=0;tt<o;tt++)
{
if(tt!=i)
array[tt]=(*tab)[tt];
else
array[tt]='6';
}
array[o]='';
*tab=array;
// printf("n%s",*tab);
}
else
{
char array[o+2];
int tt;
for(tt=0;tt<=o;tt++)
{
array[tt]='5';
}
array[o+1]='';
*tab=array;
// printf("n%s",*tab);
}
}
int main()
{
int n;
char *test;
test="555";
increase(&test);
printf("n%s",test);
return 0;
}
Okay, so increase() is meant to replace number in test with the next number containing only 5 and 6.
What I wanted to do is to change the value of test directly by using pointer to char*. Everything seems to work just fine untill it comes to displaying changed value - it simply won't show unless was asked to do it inside the
increase() function. Once I add
printf("n%s",*tab);
inside any of conditions (commented), everything works just fine (excluding showing a double result).
What causes a problem here?
555 is just a testing value, actually any number made of 5s or 6s will do the work.
c arrays string pointers char
asked Nov 12 at 1:20
Igor
193
1
Local variables in a function stop existing when the function returns
– M.M
Nov 12 at 1:24
Okay, small correction - after adding printf in one of conditions, modified string won't be shown twice if the other condition was fulfilled
– Igor
Nov 12 at 1:26
suggest replacing:int i=strlen(*tab); int o=i; while((*tab)[i]!='5' && i>=0) { i--; }with a call tostrrchr()as that function's purpose is to find a char, starting at the end of the supplied string
– user3629249
Nov 12 at 2:20
suggest definingarrayin main() and adding the array as a parameter toincrease()
– user3629249
Nov 12 at 2:22
add a comment |
#include <stdio.h>
#include <string.h>
void increase(char **tab)
{
int i=strlen(*tab);
int o=i;
while((*tab)[i]!='5' && i>=0)
{
i--;
}
if(i>=0)
{
int tt;
char array[o+1];
for(tt=0;tt<o;tt++)
{
if(tt!=i)
array[tt]=(*tab)[tt];
else
array[tt]='6';
}
array[o]='';
*tab=array;
// printf("n%s",*tab);
}
else
{
char array[o+2];
int tt;
for(tt=0;tt<=o;tt++)
{
array[tt]='5';
}
array[o+1]='';
*tab=array;
// printf("n%s",*tab);
}
}
int main()
{
int n;
char *test;
test="555";
increase(&test);
printf("n%s",test);
return 0;
}
Okay, so increase() is meant to replace number in test with the next number containing only 5 and 6.
What I wanted to do is to change the value of test directly by using pointer to char*. Everything seems to work just fine untill it comes to displaying changed value - it simply won't show unless was asked to do it inside the
increase() function. Once I add
printf("n%s",*tab);
inside any of conditions (commented), everything works just fine (excluding showing a double result).
What causes a problem here?
555 is just a testing value, actually any number made of 5s or 6s will do the work.
c arrays string pointers char
asked Nov 12 at 1:20
Igor
193
#include <stdio.h>
#include <string.h>
void increase(char **tab)
{
int i=strlen(*tab);
int o=i;
while((*tab)[i]!='5' && i>=0)
{
i--;
}
if(i>=0)
{
int tt;
char array[o+1];
for(tt=0;tt<o;tt++)
{
if(tt!=i)
array[tt]=(*tab)[tt];
else
array[tt]='6';
}
array[o]='';
*tab=array;
// printf("n%s",*tab);
}
else
{
char array[o+2];
int tt;
for(tt=0;tt<=o;tt++)
{
array[tt]='5';
}
array[o+1]='';
*tab=array;
// printf("n%s",*tab);
}
}
int main()
{
int n;
char *test;
test="555";
increase(&test);
printf("n%s",test);
return 0;
}
Okay, so increase() is meant to replace number in test with the next number containing only 5 and 6.
What I wanted to do is to change the value of test directly by using pointer to char*. Everything seems to work just fine untill it comes to displaying changed value - it simply won't show unless was asked to do it inside the
increase() function. Once I add
printf("n%s",*tab);
inside any of conditions (commented), everything works just fine (excluding showing a double result).
What causes a problem here?
555 is just a testing value, actually any number made of 5s or 6s will do the work.
c arrays string pointers char
c arrays string pointers char
asked Nov 12 at 1:20
Igor
193
asked Nov 12 at 1:20
Igor
193
asked Nov 12 at 1:20
Igor
193
asked Nov 12 at 1:20
Igor
193
asked Nov 12 at 1:20
Igor
193
193
1
Local variables in a function stop existing when the function returns
– M.M
Nov 12 at 1:24
Okay, small correction - after adding printf in one of conditions, modified string won't be shown twice if the other condition was fulfilled
– Igor
Nov 12 at 1:26
suggest replacing:int i=strlen(*tab); int o=i; while((*tab)[i]!='5' && i>=0) { i--; }with a call tostrrchr()as that function's purpose is to find a char, starting at the end of the supplied string
– user3629249
Nov 12 at 2:20
suggest definingarrayin main() and adding the array as a parameter toincrease()
– user3629249
Nov 12 at 2:22
add a comment |
1
Local variables in a function stop existing when the function returns
– M.M
Nov 12 at 1:24
Okay, small correction - after adding printf in one of conditions, modified string won't be shown twice if the other condition was fulfilled
– Igor
Nov 12 at 1:26
suggest replacing:int i=strlen(*tab); int o=i; while((*tab)[i]!='5' && i>=0) { i--; }with a call tostrrchr()as that function's purpose is to find a char, starting at the end of the supplied string
– user3629249
Nov 12 at 2:20
suggest definingarrayin main() and adding the array as a parameter toincrease()
– user3629249
Nov 12 at 2:22
1
1
Local variables in a function stop existing when the function returns
– M.M
Nov 12 at 1:24
Local variables in a function stop existing when the function returns
– M.M
Nov 12 at 1:24
Okay, small correction - after adding printf in one of conditions, modified string won't be shown twice if the other condition was fulfilled
– Igor
Nov 12 at 1:26
Okay, small correction - after adding printf in one of conditions, modified string won't be shown twice if the other condition was fulfilled
– Igor
Nov 12 at 1:26
suggest replacing:
int i=strlen(*tab); int o=i; while((*tab)[i]!='5' && i>=0) { i--; } with a call to strrchr() as that function's purpose is to find a char, starting at the end of the supplied string– user3629249
Nov 12 at 2:20
suggest replacing:
int i=strlen(*tab); int o=i; while((*tab)[i]!='5' && i>=0) { i--; } with a call to strrchr() as that function's purpose is to find a char, starting at the end of the supplied string– user3629249
Nov 12 at 2:20
suggest defining
array in main() and adding the array as a parameter to increase()– user3629249
Nov 12 at 2:22
suggest defining
array in main() and adding the array as a parameter to increase()– user3629249
Nov 12 at 2:22
add a comment |
1 Answer
1
active
oldest
votes
Lines like *tab=array; cause a problem, given that array is a local variable that goes out of scope when its enclosing block ends. In main(), test now points to memory that's invalid, and trying to use it is undefined behavior.
answered Nov 12 at 1:25
Shawn
3,4531613
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Lines like *tab=array; cause a problem, given that array is a local variable that goes out of scope when its enclosing block ends. In main(), test now points to memory that's invalid, and trying to use it is undefined behavior.
answered Nov 12 at 1:25
Shawn
3,4531613
add a comment |
Lines like *tab=array; cause a problem, given that array is a local variable that goes out of scope when its enclosing block ends. In main(), test now points to memory that's invalid, and trying to use it is undefined behavior.
answered Nov 12 at 1:25
Shawn
3,4531613
add a comment |
Lines like *tab=array; cause a problem, given that array is a local variable that goes out of scope when its enclosing block ends. In main(), test now points to memory that's invalid, and trying to use it is undefined behavior.
answered Nov 12 at 1:25
Shawn
3,4531613
Lines like *tab=array; cause a problem, given that array is a local variable that goes out of scope when its enclosing block ends. In main(), test now points to memory that's invalid, and trying to use it is undefined behavior.
answered Nov 12 at 1:25
Shawn
3,4531613
answered Nov 12 at 1:25
Shawn
3,4531613
answered Nov 12 at 1:25
Shawn
3,4531613
answered Nov 12 at 1:25
Shawn
3,4531613
3,4531613
add a comment |
add a comment |
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1
Local variables in a function stop existing when the function returns
– M.M
Nov 12 at 1:24
Okay, small correction - after adding printf in one of conditions, modified string won't be shown twice if the other condition was fulfilled
– Igor
Nov 12 at 1:26
suggest replacing:
int i=strlen(*tab); int o=i; while((*tab)[i]!='5' && i>=0) { i--; }with a call tostrrchr()as that function's purpose is to find a char, starting at the end of the supplied string– user3629249
Nov 12 at 2:20
suggest defining
arrayin main() and adding the array as a parameter toincrease()– user3629249
Nov 12 at 2:22