So the function I declared doesn't seem to working as intended, and even so, I don't think that's the proper way to compare characters by incrementing the pointer, so I'm generally lost here. To be honest, pointers have always confused me and I really need to learn how to use them if I am going to get better at C. Thanks anyone for any help!
Here's the code I have, and if this helps, the purpose of the program is for you to enter a line of text, enter a single character to search in that line of text, and then find those characters using CharIsAt. (Will add the following later) the values stored in "found" will then be printed as well.
#include <stdio.h>
#define SIZE 41
int CharIsAt(char *pStr,char ch,int loc[],int mLoc);
int main(void){
char array[SIZE],search;
int found[SIZE],chars;
printf("Enter a line of text(empty line to quit): ");
while (fgets(array,SIZE, stdin)!=NULL && array[0]!='\n')
{
printf("Enter a character to search: ");
search=getchar();
chars=CharIsAt(array,search,found,SIZE);
}
return 0;
}
int CharIsAt(char *pStr,char ch,int loc[],int mLoc){
//Searches for ch in *pStr by incrementing a pointer to access
//and compare each character in *pStr to ch.
int i,x;
for (i=0;i<mLoc;i++){
if (strcmp(pStr[i],ch)==0){
//Stores index of ch's location to loc
loc[i]=pStr[i];
x++;
}
}
//Returns the number of times ch was found
return x;
}
EDIT: Flipped sign around in the for loop. Now the program gives me a "stopped working" error.
Related
For an instance if I store ABCDE from scanf function, the later printf function gives me ABCDE as output. So what is the point of assigning the size of the string(Here 4).
#include <stdio.h>
int main() {
int c[4];
printf("Enter your name:");
scanf("%s",c);
printf("Your Name is:%s",c);
return 0;
}
I'll start with, don't use int array to store strings!
int c[4] allocates an array of 4 integers. An int is typically 4 bytes, so usually this would be 16 bytes (but might be 8 or 32 or something else on some platforms).
Then, you use this allocation first to read characters with scanf. If you enter ABCDE, it uses up 6 characters (there is an extra 0 byte at the end of the string marking the end, which needs space too), which happens to fit into the memory reserved for array of 4 integers. Now you could be really unlucky and have a platform where int has a so called "trap representation", which would cause your program to crash. But, if you are not writing the code for some very exotic device, there won't be. Now it just so happens, that this code is going to work, for the same reason memcpy is going to work: char type is special in C, and allows copying bytes to and from different types.
Same special treatment happens, when you print the int[4] array with printf using %s format. It works, because char is special.
This also demonstrates how very unsafe scanf and printf are. They happily accept c you give them, and assume it is a char array with valid size and data.
But, don't do this. If you want to store a string, use char array. Correct code for this would be:
#include <stdio.h>
int main() {
char c[16]; // fits 15 characters plus terminating 0
printf("Enter your name:");
int items = scanf("%15s",c); // note: added maximum characters
// scanf returns number of items read successfully, *always* check that!
if (items != 1) {
return 1; // exit with error, maybe add printing error message
}
printf("Your Name is: %s\n",c); // note added newline, just as an example
return 0;
}
The size of an array must be defined while declaring a C String variable because it is used to calculate how many characters are going to be stored inside the string variable and thus how much memory will be reserved for your string. If you exceed that amount the result is undefined behavior.
You have used int c , not char c . In C, a char is only 1 byte long, while a int is 4 bytes. That's why you didn't face any issues.
(Simplifying a fair amount)
When you initialize that array of length 4, C goes and finds a free spot in memory that has enough consecutive space to store 4 integers. But if you try to set c[4] to something, C will write that thing in the memory just after your array. Who knows what’s there? That might not be free, so you might be overwriting something important (generally bad). Also, if you do some stuff, and then come back, something else might’ve used that memory slot (properly) and overwritten your data, replacing it with bizarre, unrelated, and useless (to you) data.
In C language the last of the string is '\0'.
If you print with the below function, you can see the last character of the string.
scanf("%s", c); add the last character, '\0'.
So, if you use another function, getc, getch .., you should consider adding the laster character by yourself.
#include<stdio.h>
#include<string.h>
int main(){
char c[4+1]; // You should add +1 for the '\0' character.
char *p;
int len;
printf("Enter your name:");
scanf("%s", c);
len = strlen(c);
printf("Your Name is:%s (%d)\n", c, len);
p = c;
do {
printf("%x\n", *(p++));
} while((len--)+1);
return 0;
}
Enter your name:1234
Your Name is:1234 (4)
31
32
33
34
0 --> last character added by scanf("%s);
ffffffae --> garbage
Hi everyone I want check input(char or int) polindrome or not but I can't do this. Can you help me ? I have error message
"invalid conversation char to char*"
I think this is a simple problem but I couldn't solve it already. Thank you for your help.
#include <stdio.h>
#include <string.h>
#include <ctype.h>
#include <stdlib.h>
int main()
{
int r,sum=0,temp;
char a;
char *b = &a;
printf("Enter a string or number\n");
scanf("%c", &a);
if ( isalpha( a ) )
{
b = a;
strrev(b);
if (strcmp(a, b) == 0)
printf("The string is a palindrome.\n");
else
printf("The string isn't a palindrome.\n");
}
else if ( isdigit( a ) )
{
temp=a;
while(a>0)
{
r=a%10;
sum=(sum*10)+r;
a=a/10;
}
if(temp==sum)
printf("palindrome number ");
else
printf("not palindrome");
}
return 0;
}
Since you mentioned that you are a student and want to learn C, I am not going to write a code but will try to point you into the right direction.
To solve this, you need:
Get a alphanumeric string
Check if it's a palindrome (you have several options there)
First of all, to get a string, you have in your code:
char a;
scanf("%c", &a);
A hint: this is only getting you one character. To get a string, you first need to allocate an array instead of one single char, and then use scanf with a different argument, not %c.
This part of the task is completely independent of the second part. I recommend first to make sure that this part works before proceeding further. You can do it by getting a string and then immediately printing it. This way you can see what you are actually dealing with.
Once you have your string, proceed with analyzing it. You could revert it and then compare to original (that's what your code is suggesting), but it's probably easier to do something like this:
Find string length
Compare each symbol in the string with its symmetrical counterpart. For example, if string length is 10, you'll need to compare symbol #0 with symbol #9, symbol #1 with symbol #8 etc. etc. Hint: you'll need to use a loop here.
I am struggling with the concept of replacing substrings within strings. This particular exercise does not want you to use built in functions from <string.h> or <strings.h>.
Given the string made up of two lines below:
"Mr. Fay, is this going to be a battle of wits?"
"If it is," was the indifferent retort, "you have come unarmed!"
I have to replace a substring with another string.
This is what I have so far, and I'm having trouble copying the substring to a new array, and replacing the substring with the new string:
#include <stdio.h>
#include <string.h>
int dynamic();
int main()
{
char str[]="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";
int i, j=0, k=0, l=0, n=0;
unsigned int e = n-2;
char data[150];
char newData[150];
char newStr[150];
printf("Give me a substring from the string");
gets(data);
printf("Give me a substring to replace it with");
gets(newData);
dynamic();
for (i=0; str[i] != '\0'; i++)
{
if (str[i] != data[j])
{
newStr[l] = str[i];
l++;
}
else if ((str[i+e] == data[j+e]) && (j<n))
{
newStr[l] = newData[j];
j++;
l++;
e--;
}
else if ((str[i+e] == data[j+e]) && (j>=n))
{
j++;
e--;
}
else
{
newStr[l] = str[i];
l++;
}
}
printf("original string is-");
for (k=0; k<n; k++)
printf("%c",str[k]);
printf("\n");
printf("modified string is-");
for(k=0; k<n; k++)
printf("%c",newStr[k]);
printf("\n");
}
int dynamic()
{
char str[]="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";
int i, n=0;
for (i=0; str[i] != '\0'; i++)
{
n++;
}
printf("the number of characters is %d\n",n);
return (n);
}
I tried your problem and got output for my code. Here is the code-
EDIT- THIS IS THE EDITED MAIN CODE
#include <stdio.h>
#include <string.h>
int var(char *); //function declaration. I am telling CPU that I will be using this function in the later stage with one argument of type char *
int main() //main function
{
char *str="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";
int i,j=0,k=0,l=0;
char data[] = "indifferent";
char newData[] = "nonchalant";
char newStr[150];
//here 'n' is returned from the 'var' function and is received in form of r,r1,r2,r3.
int r=var(str); //getting the length of str from the function 'var' and storing in 'r'
int r1=var(data); //getting the length of data from the function 'var' and storing in 'r1'
int r2=var(newData); //getting the length of newData from the function and storing in 'r2'
unsigned int e=r1-2; //r1-2 because r1 is the data to be replaced. and string index starts from 0. Here r1 is of length 12. but we dont need to check last
//character because it is null character and the index starts from 0. not from 1. so, it is 0 to 11 and 11th is '\0'. so "12-"2"=10" characters to be compared.
for(i=0;str[i]!='\0';i++)
{
if(str[i]!=data[j])
{
newStr[l]=str[i];
l++;
}
else if((str[i+e]==data[j+e]) && (j<r2))
{
newStr[l]=newData[j];
j++;
l++;
e--;
}
else if((str[i+e]==data[j+e]) && (j>=r2))
{
j++;
e--;
}
else
{
newStr[l]=str[i];
l++;
}
}
int r3=var(newStr); //getting the length of str from the function and storing in 'r'
printf("original string is-");
for(k=0;k<r;k++)
printf("%c",str[k]);
printf("\n");
printf("modified string is-");
for(k=0;k<r3;k++)
printf("%c",newStr[k]);
printf("\n");
} // end of main function
// Below is the new function called 'var' to get the character length
//'var' is the function name and it has one parameter. I am returning integer. so, it is int var.
int var(char *stri)//common function to get length of strings and substrings
{
int i,n=0;
for(i=0;stri[i]!='\0';i++)
{
n++; //n holds the length of a string.
}
// printf("the number of characters is %d\n",n);
return (n); //returning this 'n' wherever the function is called.
}
Let me explain few parts of the code-
I have used unsigned int e, because I don't want 'e' to go negative.(I will explain more about this later).
In the first for loop, I am checking whether my string has reached the end.
In first 'IF' condn, I am checking whether the first character of string is NOT-EQUAL to the first character of the word which needs to be replaced. If condition satisfies, print regularly thr original string.
ELSE IF, i.e(first character of string is EQUAL to the first character of the word)then check the next few characters to make sure that the word matches. Here, I used 'e' because it will check the condition for str[i+e] and data[i+e]. example- ai notequalto ae. If I had not used 'e'in code,... after checking the first character itself, newdata would have been printed in newstr. I used 'e'=5 because the probabilty of 1st letter and 5th letter being the same in data and the str is less. You can use 'e'=4 also. No rule that you have to use 'e'=5 only.
Now, I am decrementing 'e' and checking whether the letters in the string is same or no. I can't increment because, there is a certain limit of size of a string. As, I used unsigned int, 'e' won't go down below 0.
ELSE, (this means that only first letter is matching, the 5th letter of str and data are not matching), print the str in newstr.
In the last FOR loop, I have used k<114 because, that much characters are there in the string. (You can write a code to find how many characters are there in a string. No need to manually count).
And lastly, I have used conditions (j<10) and (j>=10) along with ELSE-IF condition because, in first ELSE-IF, the new data is ofsize 10. So, even if the word to be replaced is more than 10,say 12 for example. I don't need the extra 2 bits to be stored in new data. So, if the size is more than 10, just bypass that in the next ELSE-IF condition. Note that this 10 is the size of new word. So, it varies if your word is smaller or bigger. And , in second ELSE-IF, I am not incrementing 'l'(l++) because, here, I am not putting anything in newstr. I am just bypassing it. So, I didn't increment.
I tried my best to put the code in words. If you have any doubt, you can ask again. I will be glad to help. And this code is NOT OPTIMAL. The numerical values used varies with the words/strings you use. Ofcourse, I can write a generalized code for that(to fetch the numerical values automatically from the strings). But, I didn't write that code here. This code works for your problem. You can change few variables like 'e' and ELSE-IF part and try to understand how the code works. Play with it.
EDIT-
include
int main()
{
char str[]="\n\"Mr. Fay, is this going to be a battle of wits?\" \n\"If it is,\" was the indifferent retort, \"you have come unarmed!\"";// I took this as string. The string which u need to calculate the length, You have to pass that as the function parameter.
int i,n=0;
for(i=0;str[i]!='\0';i++)
{
n++;
}
printf("the number of characters is %d\n",n);
return (n);
}// If you execute this as a separate program, you will get the number of characters in the string. Basically, you just have to modify this code to act as a separate function and when calling the function, you have to pass correct arguments.
//Use Pointers in the function to pass arguments.
I have written this code it is working fine for a?b?c? and a?b?c?d? but for a?b?c?d?e? it is giving one additional garbage value at the end. At the end of s there is '\0' character attached then why and how is it reading that garbage value. I tried to debug it by placing printf statements in between the code but couldn't resolve it. please help.
#include<stdio.h>
void print(char* s,char c[],int l)
{
int i,j=0;
for(i=0;s[i]!='\0';i++)
{
if(s[i]=='?')
{
printf("%c",c[j]);
j++;
}
else
printf("%c",s[i]);
}
printf(", ");
}
void permute(char *s,char c[],int l,int index)
{
if(index==l)
{
print(s,c,l);
return;
}
c[index]='0';
permute(s,c,l,index+1);
c[index]='1';
permute(s,c,l,index+1);
}
int main()
{
char s[10],c[10];
printf("Enter a string.");
scanf("%s",s);
int i,ct=0;
for(i=0;s[i]!='\0';i++)
{
if(s[i]=='?')
ct++;
}
permute(s,c,ct,0);
return 0;
}
My output was like this :-
a0b0c0d0e0♣, a0b0c0d0e1♣,
...and so on.
As we can see from your code, with an array defined like char s[10] and the input being
a?b?c?d?e?
is too big an input to be held in s along with the null-terminator by
scanf("%s",s);
You need to use a bigger array. Otherwise, in attempt to add the terminating null after the input, the access is being made to out-of-bound memory which invokes undefined behaviour.
That said, never allow unbound input to the limited-sized array, always use the field-width to limit the input length (in other words, reserve the space for null-terminator), like
scanf("%9s",s);
The code is producing the correct output here, but note that it has undefined behavior for strings of size greater than or equal to 10 chars, because that's the size of your buffer.
So, for a?b?c?d?e? you need a buffer of at least 11 characters, to account for the null terminator. You should make s bigger.
See actually in C what happens in String is that everytime it appends a '\0' character at last.
Now notice in C there is nothing called string.
It's array of characters.
So if you have defined like this-
char s[10]
This actually accepts an array of less than of 9 characters as the last one will be the '\0' character.
If you add more than 9 character it will give erroneous output.
I wanted to write a program which counts the occurrences of each letter in a string, then prints one of each letter followed by the count for that letter.
For example:
aabbcccd -
Has 2 a, 2 b, 3 c, and 1 d
So I'd like to convert and print this as:
a2b2c3d1
I wrote code (see below) to perform this count/conversion but for some reason I'm not seeing any output.
#include<stdio.h>
main()
{
char array[]="aabbcccd";
char type,*count,*cp=array;
while(cp!='\0'){
type=*cp;
cp++;
count=cp;
int c;
for(c=1;*cp==type;c++,cp++);
*count='0'+c;
}
count++;
*count='\0';
printf("%s",array);
}
Can anyone help me understand why I'm not seeing any output from printf()?
char array[]="aabbcccd";
char type,*count,*cp=array;
while(cp!='\0'){
*cp is a pointer it's pointing to the address of the start of the array, it will never be == to a char '\0' so it can't leave the loop.
You need to deference the pointer to get what it's pointing at:
while(*cp != '\0') {
...
Also, you have a ; after your for loop, skipping the contents of it:
for(c=1;*cp==type;c++,cp++); <-- this ; makes it not execute the code beneath it
After fixing both of those problems the code produces an output:
mike#linux-4puc:~> ./a.out
a1b1c2cd
Not the one you wanted yet, but that fixes your problems with "printf not functional"
Incidentally, this code has a few other major problems:
You try to write past the end of the string if the last character appears once (you write a '1' where the trailing '\0' was, and a '\0' one character beyond that.
Your code doesn't work if a character appears more than 9 times ('0' + 10 is ':').
Your code doesn't work if a character appears more than 2 times ("dddd" doesn't become "d4"; it becomes "d4dd").
Probably line-buffering. Add a \n to your printf() formatting string. Also your code is very scary, what happens if there are more than 9 of the same character in a row?
1) error correction
while(*cp!='\0'){
and not
while(cp!='\0'){
2) advice
do not use array[] to put in your result user another array to put in your rusel it's more proper and eay
I tried to solve your question quickly and this is my code:
#include <stdio.h>
#define SIZE 255
int main()
{
char input[SIZE] = "aabbcccd";/*input string*/
char output[SIZE]={'\0'};/*where output string is stored*/
char seen[SIZE]={'\0'};/*store all chars already counted*/
char *ip = input;/*input pointer=ip*/
char *op = output;/*output pointer = op*/
char *sp = seen;/*seen pointer=sp*/
char c,count;
int i,j,done;
i=0;
while(i<SIZE && input[i]!='\0')
{
c=input[i];
//don't count if already searched:
done=0;
j=0;
while(j<SIZE)
{
if(c==seen[j])
{
done=1;
break;
}
j++;
}
if(done==0)
{//if i never searched char 'c':
*sp=c;
sp++;
*sp='\0';
//count how many "c" there are into input array:
count = '0';
j=0;
while(j<SIZE)
{
if(ip[j]==c)
{
count++;
}
j++;
}
*op=c;
op++;
*op=count;
op++;
}
i++;
}
*op='\0';
printf("input: %s\n",input);
printf("output: %s\n",output);
return 0;
}
It's not a good code for several reasons(I don't check arrays size writing new elements, I could stop searches at first empty item, and so on...) but you could think about it as a "start point" and improve it. You could take a look at standard library to copy substring elements and so on(i.e. strncpy).