Print a string using putch with pointers in C - c

Question

So I'm trying to print an inputted string using putch and a little bit of pointers.
Here is my current code:
#include<stdio.h>
#include<conio.h>
#include<string.h>
void printer(char *c);
char *c;
char ch;
main(){
clrscr();
printf("Enter a string: ");
scanf("%s",&ch);
c = &ch;
printer(c);
getch();
}
void printer(char *c){
int x;
for(x=0;x<strlen(c);x++){
putch(*c);
}
}
The problem is that i can only print the first character of the string, also for some reason strlen always return 3 for strings that are 3 characters and below.
Do I have to use array for this so that I can use putch since it is limited to only 1 character output.

Answer 1

One of the problems is that your printer() function is not printing anything other than the first character. There are two ways of approaching this. Using pointers:
void printer(char const *c){
while ( *c != '\0' ) {
putch(*c);
c++;
}
}
And using pointer arithmetic:
void printer(char const *c) {
int x;
for ( x=0; x < strlen(c); x++ ) {
putch( *(c + x) );
}
}
The biggest problem is that you are attempting to store a string in a single character in memory. That's just asking for problems.
char ch;
scanf("%s",&ch); // NO NO NO NO NO
Instead declare your buffer (to store the string in) as an array big enough for the biggest string you expect:
char ch[512];
scanf("%s", ch);

Answer 2

First off, you pass a pointer to "storage for one character" to scanf. Anything that happens after that is in nsal demons territory.
Second, scanf does not allocate storage for your input, so even if you'd passed c instead of &ch, you would not be any better off.
Third, you really should be declaring your variables inside main rather than using global variables.
Something like this may be closer to what you actually want:
void output (char *c)
{
char *cp;
for (cp = c; *cp; cp++) {
putch(*cp);
}
}
int main (void)
{
char input[80];
printf("Please input a string: ");
scanf("%s\n", input);
output(input);
}

Answer 3

try this code:
#include<stdio.h>
#include<conio.h>
#include<string.h>
void printer(char *c);
char *c;
char buffer[1000];// use as a buffer
void main(){
clrscr();
printf("Enter a string: ");
scanf("%s",buffer);//read the input to the buffer
c=(char*)malloc(strlen(buffer)+1);//alloc memory with len of input + 1 byte to "\0"(end of string)
strcpy(c,buffer);//copy the input from the buffer to the new memory
printer(c);
getch();
free(c);//free the memeory
}
void printer(char *c)
{
int x;
for(x=0;x<strlen(c);x++){//move the index string pointer to next char in the string
putch(c[x]);//print the char to the screen
}
}
1)You cant use char to save a string u need char*!!!
2)You can get input to memory that not allocated!!!! because of that u must read the input to buffer after that alloc string by size of the input inside the buffer!

Answer 4

Your code print only first character because c is always pointing to first character of the array. For printing total string you need to increment character pointer as well
You need to do like this
void printer(char *c){
while(*c != '\0'){
putch(*c);
c++;
}
}

Answer 5

First calculate the length of the string & then use above implementation like this-
void printer(char *c){
int i, length;
length=strlen(c)
for(i=0;i<lenth;i++,c++){
putch(*c);
}
}
It should work I think.

Answer 6

#include<stdio.h>
#include<conio.h>
#include<string.h>
void printer(char *c);
char *c;
char ch;//the ch should be a array
main(){
clrscr();
printf("Enter a string: ");
scanf("%s",&ch);//the ch don't need '&'
c = &ch;//the ch don't need '&'
printer(c);
getch();
}
void printer(char *c){
int x;
for(x=0;x<strlen(c);x++){
putch(*c);
}
}