The program stops working.
Even if I put only one int.
I tried many different ways but can't figure out what is wrong.
I am trying to take input of integers separated by space.
There can be any no of integers.
#include<stdio.h>
#include<stdlib.h>
int main(void) {
int i,j=0;
int b[100];
char a[100];
fgets(a,100,stdin);
for(i=0;i<strlen(a);i++)
{
b[i] = atoi(a[j]);
j=j+2;
}
for(i=0;i<strlen(b);i++)
{
printf("%d ",b[i]);
}
}
Here is the prototype of atoi you have to use a character array but you are sending a character only.atoi(str[i])
int atoi(const char *str)
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int val;
char str[20];
strcpy(str, "98993489");
val = atoi(str);
printf("String value = %s, Int value = %d\n", str, val);
return(0);
}
Do the following:
for(i = 0; i < strlen(a); i += 2)
{
b[j] = a[i];
j++;
}
though atoi() accepts only a string argument or u can say constant char pointer and str[i] is nothing but a pointer pointing at a single character of the character array.
Therefore once we pass atoi(str[i]) , that means we are passing a character to the function, which will give an error.
Therefore you must pass the address of that particular character from where you want to convert the substring into a number i.e. atoi(&str[i]).
Related
I got a task from my college which is to find the length of a string using pointers in a function.Here is my code:
#include <stdio.h>
#include <stdlib.h>
#define max 100
int lengthchar(char *array1[])
{
int a, x = 0;
for (a = 0; *(array1++) != '\0'; a++)
{
x++;
}
return x;
}
int main()
{
char arr1[max];
int length;
printf("Enter string\n");
gets(arr1);
length = lengthchar(arr1);
printf("Length=%d", length);
return 0;
}
But when I give input something bellow:
Enter string: av
Length=9
It shows the wrong length, what's my mistake?
This is because of the way to pass a string pointer as an argument to the function.
Just change char *array1[] to char *array1 or array1[].
Have a look at the implementation below:
int lengthchar(char array1[])
{
int a,x=0;
for(a=0;*(array1++)!='\0';a++)
{
x++;
}
return x;
}
PS: variable a can be removed by using a while loop.
In your function signature, you are telling the compiler that lenghtchar() expects a pointer to character strings, or **char in other words.
What you really want to do is to change your function from int lengthchar(char *array1[]) to int lengthchar(char array1[]) or int lengthchar(char *array1). This is a bit tricky since in C, you can address an array by using the address of its first element (aka, by using pointer to its first item).
Expert C Programming has a dedicated chapter on this topic.
Now, coming to your lengthchar() function, I would do some refactoring to eliminate the variable a and use a while loop instead. I have also included another alternative implementation that relies on pointer arithmetic (more fun to me :) )
Note also that I used fgets() instead of gets() which is considered deprecated since it does not do any bounds checking.
#include <stdio.h>
#include <stdlib.h>
#define max 100
/*
* returns the lenght of the string excluding the terminating
* NULL character
*/
int lengthchar(char *array1) {
int x = 0;
while (*array1++)
x++;
return x-1;
}
int lengthchar1(char *array1){
char *p;
for (p = array1; *p; p++)
;
return p - array1 - 1; /* -1 for \0 */
}
int main() {
char arr1[max];
int length;
printf("Enter string\n");
fgets(arr1, max, stdin);
length = lengthchar(arr1);
printf("Length=%d", length);
return 0;
}
On declaring or defining a function that shall take an array of type T as an argument, use the notation: T *array or T array[]; T may stand for any valid C data type such as char, int, float, double, etc...
As for your loop, the variable a seems to be redundant because it has no effect on any part of the program. The return value of the function lengthchar() is one more than the number of characters inputted.
Also you should avoid the function gets() because it is deemed dangerous and has been removed from the C standard; use fgets() instead.
Your code should look something like this:
#include <stdio.h>
#include <string.h>
#define max 100
int lengthchar(char *str)
{
int len = 0;
while (*str++)
{
len++;
}
return len;
}
int main()
{
char str1[max];
printf("Enter string:");
fgets(str1, max, stdin);
str1[strcspn(str1, "\n")] = 0; // remove new-line character
int length = lengthchar(str1);
printf("Length = %d\n", length);
return 0;
}
I'm trying to find to find which position a number has in a sentence in c. I'm kinda new to programming and i don't know why my code isn't working.
i keep getting this warning but i have no clue what it means (english isn't my first language):
passing argument 1 of 'strcmp' makes pointer from integer without a cast [-Wint-conversion] Main.c /TweeIntegers line 20 C/C++ Problem
My code:
#include <stdio.h>
#include <string.h>
int main()
{
int i, y;
char x;
char text1[] = "een stuk text";
char text2[] = "k";
for ( i = 0; i < strlen(text1); i++ )
{
x = text1[i];
y = strcmp( x, text2 )
}
printf("%d", i);
return 0;
}
strcspn will search for list of characters and return the index of the first match. In this case the list is only the letter k. In case of no match found it returns the length of the string searched.
#include <stdio.h>
#include <string.h>
int main()
{
int y = 0;
char text1[] = "een stuk text";
char text2[] = "k";
y = strcspn ( text1, text2);
printf("%d", y);
return 0;
}
if you are only looking for a char and the first position then you can use the following code:
#include <stdio.h>
#include <string.h>
int main()
{
int i;
char text1[] = "een stuk text";
char charYouLookFor = 'k';
for ( i = 0; i < strlen(text1); i++ )
{
if (text1[i] == charYouLookFor)
break;
}
printf("%d", i);
return 0;
}
If you are looking for the position of a text in a text or for the second position of the char the code needs to be more complex.
You are trying to compare a single character with a string, but strcmp() compares two strings. You can solve this by dropping the whole loop and simply use strchr() to locate the character. strstr(text1, text2) would have worked as well, since text2 is a string (properly null-terminated).
Pre-made search functions in string.h:
strchr() finds a single character inside a string.
strstr() finds a sub-string inside another string.
strpbrk() finds any character from a list of specified characters inside another string.
You can only use strcmp() to compare whole strings, which is not what you're trying to do.
To locate a single character in a string, just use strchr(). No need to loop yourself:
#include <stdio.h>
#include <string.h>
int main()
{
int i, y;
char x;
const char text[] = "een stuk text";
char letter = 'k';
const char * const found = strchr(text, letter);
if(found != 0)
printf("%d\n", (int) (found - text));
return 0;
}
This prints:
7
Which is correct, it's the 8th letter.
I have I problem. I get 2 warnings from console, but I dont know what's wrong with my code. Can you have look?
Program suppose to show lines with at least 11 characters and 4 numbers
#include <stdio.h>
#include <ctype.h>
int main()
{
char line[200];
printf("Enter a string: \n");
while(fgets(line, sizeof(line),stdin))
{
int numberAlpha = 0;
int numberDigit = 0;
if(isalpha(line)) numberAlpha++;
else if(isdigit(line)) numberDigit++;
if(numberAlpha+numberDigit>10 && numberDigit>3) printf("%s \n", line);
}
return 0;
}
Both isalpha() and isdigit() takes an int, not a char *, as argument.
In your code, by passing the array name as the argument, you're essentially passing a char * (array name decays to the pointer to the first element when used as function argument), so, you're getting the warning.
You need to loop over the individual elements of line and pass them to the functions.
That said, just a suggestion, for hosted environment, int main() should be int main(void) to conform to the standard.
isalpha and isdigit are supposed to test if a char taken as int (a char can be safely converted to an int) is the encoding of an alphanumeric or digit character. You pass a char array, not an individual char. You need to test each char of the string you got, so you need a loop as:
for (int i=0; i<strlen(line); i++) {
if (isalpha(line[i])) numberAlpha++;
...
}
It is better to compute the length once:
int length = strlen(line);
for (int i=0; i<length; i++) {
...
}
You may also use a pointer to move along the string:
for (char *ptr = line; *ptr!=`\0`; ptr++) {
if (isalpha(*ptr)) ...
...
}
isalpha() and isdigit() functions take an int. But you are passing a char* i.e. the array line gets converted into a pointer to its first element (see: What is array decaying?). That's what the compiler complains about. You need to loop over line to find the number of digits and alphabets in it.
Also note that fgets() will read in the newline character if line has space. So, you need to trim it out before counting.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main(void)
{
char line[200];
printf("Enter a string: \n");
while(fgets(line, sizeof(line),stdin))
{
int numberAlpha = 0;
int numberDigit = 0;
line[strcspn(line, "\n")] = 0; // Remove the trailing newline, if any.
for (size_t i = 0; line[i]; i++) {
if(isalpha((unsigned char)line[i])) numberAlpha++;
else if((unsigned char)isdigit(line[i])) numberDigit++;
}
printf("alpha: %d, digits:%d \n", numberAlpha, numberDigit);
}
return 0;
}
Ok, i got something like this:
#include <stdio.h>
#include <ctype.h>
int main()
{
char line[200];
printf("Enter a string: \n");
while(fgets(line, sizeof(line),stdin))
{
int numberAlpha = 0;
int numberDigit = 0;
int i;
for(i=0; i<strlen(line); i++){
if(isalpha(line[i])) numberAlpha++;
else if(isdigit(line[i])) numberDigit++;
}
if(numberAlpha+numberDigit>10 && numberDigit>3) printf("%s \n", line);
}
return 0;
}
Now the question is, if it is passible to make it first accepts data and then display only those line which follows the if statment. Now it shows line just after input it.
So after a few years of inactivity after studying at uni, I'm trying to build up my c experience with a simple string reverser.
here is my code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/*
*
*/
int main(int argc, char** argv) {
reverser();
return(0);
}
int reverser(){
printf("Please enter a String: ");
//return (0);
int len;
char input[10];
scanf("%s",&input);
int quit = strcmp(input,"quit");
if(quit == 0){
printf("%s\n","Program quitting");
return(0);
}
len = strlen(input);
printf("%i\n",len);
char reversed[len];
int count = 0;
while (count <= (len-1)){
//printf("%i\n",(len-count));
reversed[count] = input[(len-1)-count];
count++;
}
//printf("%s\n",input);
printf(reversed);
printf("\n");
reverser();
}
When I input "hello", you would expect "olleh" as the response, but I get "olleh:$a ca&#",
How do I just get the string input reversed and returned?
Bombalur
Add a '\0' at the end of the array. (as in, copy only chars until you reach '\0' - which is the point at array[strlen(array)], then when you're done, add a '\0' at the next character)
Strings are conventionally terminated by a zero byte. So it should be
char reversed[len+1];
And you should clear the last byte
reversed[len] = (char)0;
you forgot the \0 at the end of the string
This is because you are creating an array with size 10. When you take in some data into it (using scanf) and the array is not filled up completely, the printf from this array will give junk values in the memory. You should iterate for the length of the input by checking \n.
must have a size + 1 to string length so that you can have a \0 at the end of string that will solve your problem
The following is a (simple and minimal implementation of) string reverse program (obviously, error conditions, corner cases, blank spaces, wider character sets, etc has not been considered).
#include <stdio.h>
int strlen(char *s)
{
char *p = s;
while (*p)
p++;
return p - s;
}
char * strrev(char a[])
{
int i, j;
char temp;
for (i=0, j=strlen(a)-1 ; i<j ; i++, j--) {
temp = a[i];
a[i] = a[j];
a[j] = temp;
}
return a;
}
int main()
{
char str[100];
printf("Enter string: ");
scanf("%s", str);
printf("The reverse is %s \n", strrev(str));
return 0;
}
Hope this helps!
I have a problem in C where i have to find number of occurrence of each character in a string.Suppose i have string like "amitamt" and output should be like "a2m2it2" .I have a routine from which i can find no of occurrence of a particular character.
int count_chars(const char* string, char ch)
{
int count = 0;
int i;
int length = strlen(string);
for (i = 0; i < length; i++)
{
if (string[i] == ch)
{
count++;
}
}
return count;
}
But I am not sure how could I count each character of string
If you have an ASCII string, create an int array of size 256. Then loop through the string and increment the value in the int array on position x. While x is the ASCII value of the character in your string you're looping through.
if i have any mistakes like syntax please excuse as im working on vb , Im unable to figure out where to put braces or brackets ,
and I belive strchr makes your task easier
#include <stdio.h>
#include <string.h>
int str_occ (char *pch ,char a)
{
int i = 0;
char *p;
p=strchr(pch,a);
while (p!=NULL)
{
i = i+1;
p = strchr(p+1,a);
}
return i;
}
To explain the code *pch is the string you have to pass ,char a is the alphabet you are searching to find how many times its occurring and int i returns the value of number of occurrences
say sample
int main()
{
char a[]="hello world";
int i;
i=str_occ(a,'l');
printf("%d",i);
}
output is 3
You can make the code as per your requirements, keep caling the function inside a loop , I mean rotate your elements