why it prints nothing in output?? please clarify me this doubt

Question

#include <stdio.h>
#include<string.h>

int main()
{
char p[20];

char *s = "string";

int length = strlen(s);

int i;

for (i = 1; i < length; i++) {

     p[i] = s[length - i];

printf("%s",p);
}
return 0;
}

Answer 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?

Answer 2

char p[20];
char *s = "string";
int length = strlen(s);
int i;
for (i = 0; i < length; i++)
p[i] = s[length -1 - i];

p[i] = '\0';
printf("%s",p);

Answer 3

becoz you did not consider that array Index starts counting from ZERO.

Answer 4

~ swim ~ I'm not changing the value of s but I'm storing it on another variable right but still why it is not modified

Answer 5

~ swim ~ since from start it was there

Answer 6

~ 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

Answer 7

codemonkey I know the code fix ... but I want to know the reason why it is printing garbage value? when p[6]

Answer 8

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;
}

Answer 9

~ swim ~ I want to know why it prints nothing... I don't want code fix

Answer 10

~ swim ~ but instead of char *s I'm giving char s[] but now it is printing in output.. how it works with char s[] and not working with char *s?

Answer 11

~ swim ~ so char *s cannot be copied to any other variables??

Answer 12

~ 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;
}

Answer 13

codemonkey yes we can copy but I'm not modifying *s ... I'm storing s[0] to p[0] but why it prints nothing??

I don't want to fix the code.... I want to know the reason why it prints nothing

Answer 14

codemonkey here also I'm not copying null but it's not printing 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;
}

Answer 15

~ 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

Answer 16

~ swim ~ okay see the below scenario
..
I'm not storing anything in p[0] ... but it is printed as tring ... please explain this

you told if p[0] is 0 or null then it will not print anything for p

int main()
{
char p[7];

char *s = "string";

for (int i=1;i<7;i++)

{

p[i]=s[i];

}
printf("%s\n",p);

return 0;
}

Answer 17

As you use %s which is used for whole string, u can use %c as you are printing each character
Or
After the for loop:
End the string p[i] ='\0' ;
then print the string using %s

Answer 18

Hola buenas tengo un proyecto para hacer y pues me decidí por investigar una app y lo malo es que no se de progracion ni lenguajes para poder investigar el como fue su código fuente y desarrollo del código alguien que me pueda ayudar

Answer 19

As string is collection of characters, we use %s specifier to print whole string, but u r printing each character in string. So it not through the error