I get you are not modifying.
You are starting at index 1, so while s[length - i] gives you the correct character from the string pointed to by s, the code writes s[length - i] character at p[1] (which should have been p[0])
And you are not understanding that in array p if you have 0 or '\0' at p[0] then you will not get the output because the string has finished at p[0] itself.
Start with i = 0 in for loop or write
p[i-1] = s[length - i];
and your loop terminating condition should be i <= length, not i < length ( if you start from i=1)
if starting from i = 0, then i < length is write but
s[length - i] should now be s[length-i-1]
why it is not working with *s ?? please clarify me and what is difference between char []s and char *s?
am not modify I'm Just storing it to another array and printing then how it crashes?
char *s = string is a special case. The actual type of s is (const char*) which means the string pointed to by "s" cannot be modified.
You can copy contents of s char by char to an array or to another char* (dynamic array) or you can use strcpy to copy the data.
so
char *s = "string";
s[0] = 'c'; // error and crash
char p[20];
p[0] = s[0]; // fine
~ swim ~ in my case I'm not modifying s I'm reversing and storing on p only .. but why not printing
p[i]=s[length-i] ....I'm not modifying s ... I'm storing in p
It's by chance only. Your array must be having 0 at p[5] here.
Do one thing try this
After declaring char p[20]; print the values it contains.
for (int i=0;i<6;i++)
printf("%d\n", p[i]); (int used for seeing the character value as int, since some characters are unprintable. See what you get at p[5]
codemonkey
but I have given %s only...now it not giving garbage why
#include <stdio.h>
#include<string.h>
int main()
{
char p[6];
char *s = "string";
for (int i=0;i<5;i++)
{
p[i]=s[i];
}
printf("%s\n",p);
return 0;
}
You are circling around the same point.
A string declared like
char *s = "string" is readonly
where as a string declared as
char s[] = "string";
is modifiable. The compiler copies the string "string" into char array to allow modification.
I am sorry i didn't read the code properly. I am deleting my comment, since it is wrong in the context. See rodwynnejones answer. His answer is right. Sorry for the inconvenience.
You are starting from i = 1 and it is likely that p[0] has a value of null terminator ('\0') at sololearn, so the string ends at p[0] itself.
you should start the loop from 0, or write p[i-1] if starting from 1
also s[length - i] should be s[length-i-1] since the last character is at the position s[length - 1];
Then once the loop is over you should explicitly add a null terminator
s[length] = '\0'; // an array is not a string till the null terminator is added.
There is no char s[] in your code, there is char p[20] in your code.
You are iterating over the string (read only) pointed by "s" from the last to the first character and copying each character in p[i] (starting from 1st position onwards).
And as i told earlier
declaring a string like
char *s = "string"; // s is pointing to a readonly string which cannot be modified. In this
char s[] = "string"; string "string" is copied into array (including the hidden null terminator, so that it can be modified.
~ swim ~ please check the below code char *s can be copied to another variable but I don't know why it is putting some garbage value at last
int main()
{
char p[6];
char *s = "string";
for (int i=0;i<6;i++)
{
p[i]=s[i];
}
printf("%s\n",p);
return 0;
}
