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Exit status -1

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This is my code and I am not getting why is it showing exit status -1.
Please answer:)
int main()
int i;
char *c,d,*e;
scanf("%s",c); //to take in the 1st string
scanf("%c",&d);//after string the newline char needs to get in the buffer
scanf("%s",e); //to take in another string
printf("%s\n%s",c,e); //to print both the strings
return 0;


Answered by mwilki7 [earned 5 cycles]
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%s is only for strings
char c,d,e
you've got 3 chars

try instead:

char c[20]; char d[20]; char e[20];

and use scanf like so:

scanf("%s", c); scanf("%s", d); scanf("%s", e);

ampersand(&) is only for non-arrays like char, int, float
but not char[], int[], float[] etc...

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Sorry I dont know about repl putting of code, I just pasted my code down and that removed * before the c and e thus I am pasting the pic of it as well.
Sorry for that mistake so *c is a character pointer let me know why that didnt work.

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I see, repl formatted your asterisks out.
Try one backwards tick mark ` for inline code, or three for code with multiple lines.

As for your code, when you declare a char pointer:
char *c;

You need space to store stuff in it.
Arrays do this automatically because you tell it from the start how much space it should have like so:
char c[20]; // space for 20 chars already made at compile

To do this with pointers:

char *c; // declare pointer c = (char*)malloc(sizeof(char) * 20); // make space for 20 chars for this pointer scanf("%s", c); // now I can use it, just make sure to limit your input to 20 or // whatever you decide to change 20 to

OR you can steal the space made by an array if you don't want to use malloc

like so:

char *c; char arr[20]; c = arr; // use arr's space for c scanf("%s", c); // read to c
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Here is a sample program that does this:

Program (allocate with array sharing)

#include <stdio.h> int main() { char *c; // declare pointer char arr[20]; // declare array c = arr; // array shares space with c scanf("%s", c);// write stuff to c printf("c: %s, arr: %s",c,arr); // print results return 0; }




c: test, arr: test

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Sankyu(Thanks) Sir for the reply, just let me know one more thing
Like why this works for a single string.
and how does it work:
I thought in the manner that being a character pointer
the pointer would store 1 character as I am entering and then will move to next byte for latter entry and as soon as I press enter it puts a '\0' at the end and completes the string,

And I also don't know like why this should be a problem even
is it because that the pointer has some garbage address stored in it and it is not a good programming practice to store our string at some garbage place though, the memory chunks might be continuous (as the char pointer
does that bit by default.)

//Please answer:) #include<stdio.h> #include<string.h> int main() { int i; char *c,d,*e; scanf("%s",c); //to take in the 1st string //scanf("%c",&d);//after string the newline char needs to get in the buffer //scanf("%s",e); //to take in another string printf("%s\n",c); //to print both the strings return 0; }

let me know is it some compiler dependant thing because I tried the same thing in Codeblocks and it didn't work.

Thanks once again for the help.

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The only thing that's allocated for pointers up front:
char *c; is the address for the variable 'c' itself.

What 'c' is currently pointing at...who knows. For the sake of deterministic behavior, it may be a good idea to initialize it to NULL

Don't do:

char *c = NULL; scanf("%s", c); // bad! writing to NULL


char *c = NULL; c = (char*)malloc(sizeof(char) * 20); // make space first scanf("%s", c); // now I can write to 'c'
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Thank you for help I will try to be a good programmer with this practice.

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Pls help me for this error