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Write a function common_char that takes two strings as arguments and returns a new string that contains a single copy of all characters that appear in either of the two strings.
For example, string1: hello; string2: world; the new string is : hellowrd (o and l were already in array from hello).
May use string function here.In other words, all characters in string1 are copied into the new string, but characters in string 2 are copied only characters that are not in string1. That is past exam question and the university did not provide answer. Here is my code.
#include <stdio.h>
#include <string.h>
char *common_char(char *string1, char *string2) {
int str_length1 = strlen(string1);
int str_length2 = strlen(string2);
char *new_string = malloc(str_length1+str_length2+1);
for (int index_1 = 0; index_1 < str_length1; index_1++) {
for (int index_2 = 0; index_2 < str_length2; index_2++) {
if (string1[index_1] == string2[index_2]) {
}
}
}
}
int main(void) {
return 0;
}
My idea is to find duplicate characters in string 2 and string 1 according to the nested loop, but there is a problem with the conditional statement, there is red line, also how to copy the character of the non-duplicate string? I know strcopy(), but how to remove the repeated characters?
I've come up with a solution that uses dynamic memory and resizes the result char* each time a new char must be added. There are two loops, the first iterates the b string and the second loop checks that non of char of the b string is repeated in the a string, if it is not repeated, then adds it. Hope you understand the realloc to resize dynamically the char* each time it must be added an element.
Firstly I initialize the result string to the size of string a so it can be all copied inside. The ordering method I think it is called bubble method.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char* common_char(char* a, char* b) {
char* result = (char*)malloc(sizeof(char)*strlen(a)+1);
int i = 0;
int j = 0;
int repeated = 0;
strcpy(result,a);
for(i=0; i<strlen(b); i++) {
for(j=0; j<strlen(result); j++) {
if(b[i] == a[j]) {
repeated = 1;
}
}
if(!repeated) {
result = (char*)realloc(result,strlen(result)+sizeof(char));
result[strlen(result)] = b[i];
result[strlen(result)+1] = '\0';
}
repeated = 0;
}
return result;
}
int main()
{
char a[] = "hello";
char b[] = "world";
char* result = common_char(a,b);
printf("%s", result);
return 0;
}
EDIT: I've modified the code to make it function. About the comment of memory allocation, I've modified the declaration of result to give it space for the '\0'. When doing the realloc, I've already considered that the realloc does not increment the strlen() because strlen() is a counter till the '\0' not of the size of the variable.
I am really new to C and in my first half year at university. This is my first questio on StackOverflow.
My task is to program it so every string stored in numbers is being converted into a decimal, without changing anything outside the main function.
I am now trying for the past 4 hours to solve this problem, where I want to iterate trough every char in the string I am currently to then, based on there position in comparison to the length to convert it into a decimal.
My only question here is to someone help me to understand how I can get the string length without using strlen() due to the fact I can't add #include <string.h>
This is what I got so far (getting the length of the array to iterate through every index):
#include <stdio.h>
#include <math.h> // Kompilieren mit -lm : gcc -Wall -std=c11 dateiname.c -lm
int main() {
char* numbers[] = {
"01001001",
"00101010",
"010100111001",
"011111110100101010010111",
"0001010110011010101111101111010101110110",
"01011100110000001101"};
// Add here..
int length = sizeof(numbers);
for ( int i = 0; i < length; i++ ){
//how do i get the string size without strlen() D:
}
return 0;
}
In C, strings are really just char arrays with a special terminator character to mark the end of the string. So, say you have something like:
char *str = "hello";
This is essentially equivalent to this:
char str[] = {'h', 'e', 'l', 'l', 'o', '\0'};
Notice that \0 character at the end of the array? This is the special terminator character that C places at the end of strings. Functions like strlen() pretty much iterate through the char array looking for the first occurrence of the \0 character and then stopping.
So, you can make your own version of strlen(), say my_strlen() like this:
int my_strlen(char *str)
{
/* Initialize len to 0 */
int len = 0;
/* Iterate through str, increment len, and stop when we reach '\0' */
while(str[len] != '\0')
len++;
/* Return the len */
return len;
}
Then within your for loop, you can just call this function. Also, note that your calculation of the size of the numbers array:
int length = sizeof(numbers);
will not give you the number of elements in the array. That code gives you the size (in bytes) or numbers which is an array of char pointers. If you want to get the number of elements, you have to divide that size by the size (in bytes) of a single element (i.e., a char pointer). So, something like this would work:
int length = sizeof(numbers) / sizeof(numbers[0]);
Your final code can look something like this:
#include <stdio.h>
#include <math.h> // Kompilieren mit -lm : gcc -Wall -std=c11 dateiname.c -lm
int my_strlen(char *str) {
/* Initialize len to 0 */
int len = 0;
/* Iterate through str, increment len, and stop when we reach '\0' */
while(str[len] != '\0')
len++;
/* Return the len */
return len;
}
int main() {
char* numbers[] = {
"01001001",
"00101010",
"010100111001",
"011111110100101010010111",
"0001010110011010101111101111010101110110",
"01011100110000001101"};
// Add here..
// Notice the change here
int length = sizeof(numbers) / sizeof(numbers[0]);
for(int i = 0; i < length; i++ ){
int str_len = my_strlen(numbers[i]);
// Do what you need with str_len
}
return 0;
}
This project can be done without computing the length of the strings. How? In C, all strings are nul-terminated containing the nul-character '\0' (with ASCII value 0) after the last character that makes up the string. When you need to iterate over a string, you just loop until the character values is 0 (e.g. the nul-character)
This is how all string function know when to stop reading characters. Since you have an array-of-pointers that contains your strings, you just need to loop over each pointer and for each pointer, loop over each character until the nul-character is found.
Putting it altogether, (and noting you don't need math.h), you can do:
#include <stdio.h>
#include <math.h> // Kompilieren mit -lm : gcc -Wall -std=c11 dateiname.c -lm
int main() {
char* numbers[] = {
"01001001",
"00101010",
"010100111001",
"011111110100101010010111",
"0001010110011010101111101111010101110110",
"01011100110000001101"};
int nnumbers = sizeof numbers / sizeof *numbers; /* no. of elements */
for (int i = 0; i < nnumbers; i++) {
long long unsigned number = 0;
/* you don't care about the length, strings are nul-terminated,
* just loop until \0 is found.
*/
for (int j = 0; numbers[i][j]; j++) {
number <<= 1; /* shift left */
number += numbers[i][j] == '1' ? 1 : 0; /* add bit */
}
printf ("%s = %llu\n", numbers[i], number); /* output result */
}
return 0;
}
(note: you must use a 64-bit type to hold the converted values as "1010110011010101111101111010101110110" requires a minimum of 38 bits to represent)
Example Use/Output
Simple example output converting each string to a numeric value:
$ ./bin/binstr2num
01001001 = 73
00101010 = 42
010100111001 = 1337
011111110100101010010111 = 8342167
0001010110011010101111101111010101110110 = 92790519158
01011100110000001101 = 379917
#include <stdio.h>
int main(){
char arr[20]="Hello";
int count=0;
while(arr[count]!='\0'){
count++;
}
printf("%d",count);
return 0;
}
Look at this small code, you will understand. In C a string ended with a NULL character. We can use that advantage.
There are a few ways to do it. IMO, a simple, reasonable way to implement strlen is:
size_t string_length(const char *s) { return strchr(s, '\0') - s; }
but if you're not allowed to use strlen then you're probably not allowed to use strchr either. So you just have to count. The most idiomatic way to do that is probably a bit obscure for a complete beginner, so here is a more verbose method.
Note that your computation of the number of elements in the array is invalid, and has been corrected below.
#include <stdio.h>
int
length(const char *s)
{
int len = 0;
while( *s++ ){
len += 1;
}
return len;
}
int
main(void)
{
char *numbers[] = {
"01001001",
"00101010",
"010100111001",
"011111110100101010010111",
"0001010110011010101111101111010101110110",
"01011100110000001101"
};
int count = sizeof numbers / sizeof *numbers; /* Number of entries */
for( int i = 0; i < count; i++ ){
printf(" length of %s is %d\n", numbers[i], length(numbers[i]));
}
return 0;
}
It's pretty subjective, but IMO a more idiomatic way to write this is:
#include <stdio.h>
int
length(const char *e)
{
const char *s = e;
while( *e++ )
;
return e - s - 1;
}
int
main(void)
{
char *numbers[] = {
"01001001",
"00101010",
"010100111001",
"011111110100101010010111",
"0001010110011010101111101111010101110110",
"01011100110000001101"
};
char **e = numbers + sizeof numbers / sizeof *numbers;
for( char **t = numbers; t < e; t++ ){
printf(" length of %s is %d\n", *t, length(*t));
}
return 0;
}
Currently i'm trying to sort a paragraph based on first char of each word. I'm initializing two alphabetic (Upper and Lower) char arrays to compare to the first char of the word to the array index.
My Plan for the problem;
Initialize two Arrays with the Alphabet to know which alphabet goes before which.
Have a function that actually compares first char to the two Arrays.
return the words sorted :)
My problem is i'm trying to see if the two arrays have 26 characters to make sure every alphabet is counted for but it prints 1 and i have no idea why.
Code:
#include <stdio.h>
int main(int argc, const char * argv[]) {
char _alphabetUpperCase[100] = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R,","S","T","U","V","W","X","Y","Z"};
char _alpabetLowerCase[100] = {"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
int i = 0;
int n = 0;
while(_alpabetLowerCase[i] != NULL)
{
n = n+1;
i++;
}
printf("%d\n",n);
return 0;
}
This "A" is not same as 'A'. You should assign 'A' to a character instead.
Replace all "" strings with corresponding '' characters. e.g.
char _alpabetLowerCase[100] = {'a','b','c' ... };
You want characters or strings? "a" is a string while 'a' is a character. From your definition of array it seems you want characters. So change your code for that as
char _alpabetLowerCase[100] = {"a", ...
Instead use
char _alpabetLowerCase[100] = {'a', 'b', ....
Try this:
#include <stdio.h>
int main(int argc, const char * argv[]) {
char _alphabetUpperCase[27] = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R,','S','T','U','V','W','X','Y','Z', '\0'};
char _alphabetLowerCase[27] = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z', '\0'};
int i = 0;
int n = 0;
while(_alphabetLowerCase[i] != '\0')
{
n = n+1;
i++;
}
printf("%d\n",n);
return 0;
}
Explanation: 'a' is char, but "a" is string. Also I added 27-th character to become sure that this loop will stop after 'z' (so I changed the array size to 27 (100 was correct too, but it is wrong to write arbitrary numbers when you can put correct value)).
What's the best way to concatenate unsigned char arrays in C? Furthermore, is there a way to concatenate unsigned char arrays with char arrays? 2 of these unsigned char arrays are really just strings, but for simplicity, I'm treating them as unsigned char arrays.
The requirement is complex: there is a function that will take 1 (one) unsigned char array. That one unsigned char array is really 4 variables concatenated to make up that 1 unsigned char array. To add to the complexity, the first unsigned char array is really just a string of variable length, but its max length is 60 (i.e. sometimes it would have length = 15, other times = 60).
someFunctionAssignsFirst(unsigned char *first)
{
//it could be 15 or 60 chars long.
...
}
unsigned char first[60] = //someFunctionAssignsFirst() //This is a string i.e. "variable size string max size 60"
unsigned char second[8] = "always8."; //This is a string i.e. "01234567"
unsigned char third[32] = "always32"; //This is a cryptographic key
unsigned char fourth[32] = "always32"; //This is a cryptographic key
How would I go about getting:
unsigned char allstrings[sizeof(first)+sizeof(second)+sizeof(third)+sizeof(fourth)] = //all strings combined
?
I attempted some for loops, but the variable length first is disrupting the concatenation, and I'm sure there has to be a better way.
Full Disclosure: I'm not an expert, and I don't necessarily love C. Also for the requirement, not allowed C++ or any other language.
This is what I was trying to do, and (for clarification) I don't get a null character at the end so it's not really a string.
unsigned char *first = "this is a sample string, human readable";
unsigned char *second = "12345678" //always a number
//unsigned char third -> I have the value from before and it's a key
//unsigned char fourth -> I have the value from before and it's a key
unsigned char allstrings[sizeof(first) + sizeof(second) + sizeof(third) + sizeof(fourth)];
int counter = 0;
for (int i = 0; i <= sizeof(first); i++)
{
allstrings[counter] = first[i];
counter++;
}
for (int i = 0; i <= sizeof(second); i++)
{
allstrings[counter] = second[i];
counter++;
}
for (int i = 0; i <= sizeof(third); i++)
{
allstrings[counter] = third[i];
counter++;
}
for (int i = 0; i <= sizeof(fourth); i++)
{
allstrings[counter] = fourth[i];
counter++;
}
The allstrings variable, doesn't get anything beyond "readable" in my example above.
You need to use strcpy to copy over the first part, which is a string, then use memcpy to copy over the other 3, which are not strings but char arrays.
Note that the result is not a string but a char array, i.e. it is not null terminated.
unsigned char allstrings[strlen(first)+sizeof(second)+sizeof(third)+sizeof(fourth)];
strcpy(allstrings,first);
memcpy(allstrings+strlen(first),second,sizeof(second));
memcpy(allstrings+strlen(first)+sizeof(second),third,sizeof(third));
memcpy(allstrings+strlen(first)+sizeof(second)+sizeof(third),fourth,sizeof(fourth));
I guess you want to treat the array as buffer.
So it's fine to have the declarations,
but you don't need to define the content for this moment:
unsigned char first[60];
unsigned char second[8];
unsigned char third[32];
unsigned char fourth[32];
#define ALLSTRLEN sizeof(first) + sizeof(second) + sizeof(third) + sizeof(fourth)
unsigned char allstrings[ALLSTRLEN];
The code will keep the fixed size of arrays. and please notice that the arrays should be global or static for safety reasons.
Then you can copy the contents to arrays. I just put your code under main() to concatenate these arrays:
int main()
{
strcpy((char *)first, "this is a sample string, human readable");
// do something for second, third, fourth....
//
int counter = 0;
// first array is a normal string, we have to copy null character for it
for (int i = 0; i <= strlen((char *)first)+1; i++)
{
allstrings[counter] = first[i];
counter++;
}
for (int i = 0; i <= sizeof(second); i++)
{
allstrings[counter] = second[i];
counter++;
}
for (int i = 0; i <= sizeof(third); i++)
{
allstrings[counter] = third[i];
counter++;
}
for (int i = 0; i <= sizeof(fourth); i++)
{
allstrings[counter] = fourth[i];
counter++;
}
// allstrings is finished
}
Please notice this example just works in main() function; if you call a function to concatenate four arrays, the compiler has to pass the arrays as pointers, and the sizeof() will be wrong (equal to the pointer's size).
You can test the size by doing this:
printf("sizeof(second)=%d\n", sizeof(second));
AFunc changes what was sent to it, and the printf() outputs the changes:
void AFunc ( char *myStr, int *myNum )
{
*myStr = 's';
*myNum = 9;
}
int main ( int argc, char *argv[] )
{
char someString = 'm';
int n = 6;
AFunc(&someString, &n);
printf("%c" "%d", someString, n);
}
But what if the string was more than one char? How would the code look differently? Thanks for any help.
If it were a "string" instead of a char, you would do something like this:
#include <stdio.h>
void AFunc (char *myStr, int *myNum) {
myStr[0] = 'p'; // or replace the lot with strcpy(myStr, "pax");
myStr[1] = 'a';
myStr[2] = 'x';
myStr[3] = '\0';
*myNum = 9;
}
int main (void) {
char someString[4];
int n = 6;
AFunc(someString, &n);
printf("%s %d", someString, n);
return 0;
}
which outputs:
pax 9
A "string" in C is really an array of characters terminated by the \0 (NUL) character.
What the above code does is to pass in the address of the first character in that array and the function populates the four characters starting from there.
In C, a pointer to char isn't necessarily a string. In other words, just because you have char *x;, it doesn't mean that x is a string.
To be a string, x must point to a suitably allocated region which has a 0 in it somewhere. The data from the first character that x points to and up to the 0 is a string. Here are some examples of strings in C:
char x[5] = {0}; /* string of length 0 */
char x[] = "hello"; /* string of length 5, the array length being 6 */
char *x = "hello"; /* string of length 5. x is a pointer to a read-only buffer of 6 chars */
char *x = malloc(10);
if (x != NULL) {
strcpy(x, "hello"); /* x is now a string of length 5. x points
to 10 chars of useful memory */
}
The following are not strings:
char x[5] = "hello"; /* no terminating 0 */
char y = 1;
char *x = &y; /* no terminating 0 */
So now in your code, AFunc's first parameter, even though is a char * isn't necessarily a string. In fact, in your example, it isn't, since it only points to a memory that has one useful element, and that's not zero.
Depending upon how you want to change the string, and how the string was created, there are several options.
For example, if the myStr points to a writable memory, you could do something like this:
/* modify the data pointed to by 'data' of length 'len' */
void modify_in_place(char *data, size_t len)
{
size_t i;
for (i=0; i < len; ++i)
data[i] = 42 + i;
}
Another slightly different way would be for the function to modify data until it sees the terminating 0:
void modify_in_place2(char *data)
{
size_t i;
for (i=0; data[i]; ++i)
data[i] = 42 + i;
}
You are only dealing with chars and char pointers. None of the char pointers are valid strings as they are not null terminated.
Try defining a string and see what it looks like.
But what if the string was more than one char? How would the code look
differently? Thanks for any help
Ofcourse, you would modify the other characters as well, but in the exact same way you did the first time.
Declare a char array and pass its address
Modify values at those address
A char array would be a more clear term for a string.