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.
Related
I am trying to write a simple piece of code that merges two strings together in even and odd indexes.
This is the code
void two_strings(char a[], char b[]) {
int counta = 0, countb = 0;
int lena = strlen(a);
int lenb = strlen(b);
int lenab = lena + lenb;
char ans[lenab];
for(int i = 0; i<strlen(ans); i++) {
if(i%2 == 0) {
ans[i] = a[counta];
counta++;
}
else {
ans[i] = b[countb];
countb++;
}
}
printf("%s\n", ans);
}
This is the main:
int main() {
char a[] = "hello";
char b[] = "bye";
two_strings(a, b);
return 0;
}
I have compiled it with -Wall and didn't get any warnings or errors, and I have tried it also with long instead of int just to check if that was the issue. when I run the code it doesn't get past the first strlen(a)
Strings in C are defined as a sequence of non-null bytes followed by a terminating null byte ('\0').
The following are equivalent strings
char one[] = "hello";
char two[] = { 'h', 'e', 'l', 'l', 'o', '\0' };
Both of these would have a string length of 5, and occupy 6 bytes of memory.
String handling functions in the Standard Library expect this null-terminating byte, and you will invoke Undefined Behavior by passing a non-null-terminated array to them. strlen and printf (with %s) are examples of these kinds of functions.
In your two_strings function you are not allocating enough memory to store the null-terminating byte. You also make no attempt to place this null-terminating byte in the array.
Allocate an additional byte for the null-terminating byte, and do not attempt to take the string length of an uninitialized array.
void two_strings(char a[], char b[]) {
/* ... */
size_t length = strlen(a) + strlen(b);
char string[length + 1];
for (size_t i = 0; i < length; i++) {
/* ... */
}
string[length] = '\0';
/* ... */
}
Also note that size_t is the correct type to use when dealing with memory indexing, and is the type returned by strlen.
As for your algorithm, in the event where your input strings differ in length you will attempt to continue indexing one of the strings after you have already reached its end.
You will either want to: only take the smaller string's length of characters from the larger string, stopping when the smaller string has been exhausted; or, append the remaining characters of the larger string to the result after the smaller string has been exhausted.
A quick example of the second approach:
#include <stdio.h>
#include <string.h>
void zip_strings(const char *a, const char *b) {
size_t combined_length = strlen(a) + strlen(b);
char joined_string[combined_length + 1];
for (size_t i = 0; i < combined_length; i++) {
const char **src = i & 1 ? &b : &a;
if (!**src)
src = &a;
if (!**src)
src = &b;
joined_string[i] = *((*src)++);
}
joined_string[combined_length] = '\0';
puts(joined_string);
}
int main(int argc, char **argv) {
if (argc > 2)
zip_strings(argv[1], argv[2]);
}
./a.out hello computer
hceolmlpouter
I have a char string containing hexadecimal characters (without 0x or \x):
char *A = "0a0b0c";
from which I want to obtain
const char *B = "\x0a\x0b\x0c";
Is there an efficient way to do this? Thanks!
EDIT: To be clear, I want the resultant string to contain the 3 characters \x0a, \x0b, \x0c, not a 12 character string that says "\x0a\x0b\x0c" where the \ and x are read as individual characters.
This is what I have tried:
const char *B[12];
for (j = 0; j < 4; ++j) {
B[4 * j + 0] = '\\';
B[4 * j + 1] = 'x';
B[4 * j + 2] = A[2 * j];
B[4 * j + 3] = A[2 * j + 1];
};
B[12] = '\0';
which gives me a 12 character string "\x0a\x0b\x0c", but I want B to be as if it was assigned thus:
const char *B = "\x0a\x0b\x0c";
There are multiple confusions in your code:
the input string has 6 characters and a null terminator
the output string should be defined as const char B[3]; or possibly const char B[4]; if you intend to set a null terminator after the 3 converted bytes.
the definition const char *B[12]; in your code defines an array of 12 pointers to strings, which is a very different beast.
The for is fine, but it does not do what you want at all. You want to convert the hexadecimal encoded values to byte values, not insert extra \ and x characters.
the trailing ; after the } is useless
you set a null terminator at B[12], which is beyond the end of B.
Here is a corrected version using sscanf:
const char *A = "0a0b0c";
const char B[4] = { 0 };
for (j = 0; j < 3; j++) {
sscanf(&A[j * 2], "%2hhx", (unsigned char *)&B[j]);
}
The conversion format %2hhx means convert at most the first 2 bytes at A[j * 2] as an unsigned integer encoded in hexadecimal and store the resulting value into the unsigned char at B[j]. The cast is only necessary to avoid a compiler warning.
You can write a function that would sprintf the desired into a string, and then concat that with the destination string.
Something along these lines...
#include <stdio.h>
#include <string.h>
void createB (char B[10], const char *start)
{
char temp[10];
sprintf(temp, "\\x%c%c", start[0], start[1]);
strcat(B, temp);
}
int main ()
{
char A[] = "0a0b0c";
char B[10] = {'\0'};
for (int i=0; A[i] != '\0'; i = i+2)
{
createB(B, A+i);
}
printf("%s\n", B);
return 0;
}
$ ./main.out
\x0a\x0b\x0c
You can modify that to suit your needs or make it more efficient as you feel.
Please make edits as you please; to make it safer with necessary checks. I have just provided a working logic.
If you simply want to add "\x" before each '0' in the string-literal A with the result in a new string B, a simple and direct loop is all that is required, and storage in B sufficient to handle the addition for "\x" for each '0' in A.
For example:
#include <stdio.h>
#define MAXC 32
int main (void) {
char *A = "0a0b0c",
*pa = A,
B[MAXC],
*pb = B;
do { /* loop over all chars in A */
if (*pa && *pa == '0') { /* if chars remain && char is '0' */
*pb++ = '\\'; /* write '\' to B, adv ptr */
*pb++ = 'x'; /* write 'x' to B, adv ptr */
}
*pb++ = *pa; /* write char from A, adv ptr */
} while (*pa++); /* while chars remain (writes nul-termining char) */
puts (B); /* output result */
}
You cannot simply change A to an array with char A[] = 0a0b0c"; and then write back to A as there would be insufficient space in A to handle the character addition. You can always declare A large enough and then shift the characters to the right by two for each addition of "\x", but it makes more sense just to write the results to a new string.
Example Use/Output
$ ./bin/straddescx
\x0a\x0b\x0c
If you need something different, let me know and I'm happy to help further. This is probably one of the more direct ways to handle the addition of the character sequence you want.
#include <stdio.h>
int main(void)
{
char str1[] = "0a0b0c";
char str2[1000];
int i, j;
i = j = 0;
printf("sizeof str1 is %d.\n", sizeof(str1)-1);
for(i = 0; i < sizeof(str1)-1; i += 2)
{
str2[j] = '\\';
str2[j+1] = 'x';
str2[j+2] = str1[i];
str2[j+3] = str1[i+1];
j+=4;
}
str2[j] = '\0';
printf("%s\n", str2);
return 0;
}
I think you can do like this.
Assuming no bad input, assuming 'a' to 'f' are sequentially in order, assuming no uppercase:
// remember to #include <ctype.h>
char *input = "0a0b0c";
char *p = input;
while (*p) {
v = (isdigit((unsigned char)*p) ? *p-'0' : *p-'a'+10) * 16;
p++;
v += isdigit((unsigned char)*p) ? *p-'0' : *p-'a'+10;
p++;
printf("0x%d", v); // use v
}
While using char A[] = "0a0b0c";, as proposed by kiran, would make it possible to change the string, it wil not yet allow to insert characters. Because that would make the string longer and hence not fit into the available memory. This in turn is a problem, if you cannot create the target string right away with the needed size.
You could know the needed size in advance, if the input is always of the same length and always requires the same number of inserted characters, e.g. if like in your example, the target string is double the size of the input string. For a simple character array definition, you would need to know the size already at compile time.
char A[7] = "0a0b0c"; /* not 6, because size for the termianting \0 is needed */
char B[13] = ""; /* 2*6+1 */
So you can stay with char *A = "0a0b0c"; and make your life easier by setting up memory of appropriate size to serve as target. For that you need to first determine the length of the needed memory, then allocate it.
Determining the size if easy, if you know that it will be twice the input size.
/* inside a function, this does not work as a variable definition */
int iLengthB = 2*length(A);
char* B = malloc(iLengthB+1); /* mind the terminator */
Then loop over A, copying each two characters to B, prepending them with the two characters "\x". I assume that this part is obvious to you. Otherwise please show how you setup the program as described above and make a loop outputting each character from A separatly. Then, after you demonstrated that effort, I can help more.
I am new in C programming. I would like to ask how I can store a char in a char array? The reason why I want to do it is because I get a char from a computation I make. This computation returns a char consisting of 4. I want to look through every character of the 4 and check if they are "0". If not, then I will increment by one. This is an example of a substitution: "0xf3"- I am trying to compute the Hamming Weight of a substitution. This is how I try to count how many non-zero the char consists of:
#include <stdio.h>
int main()
{
char x = "0xf3";
char s[10] = x;
int lingh = strlen(s);
for (int i = 0; i<lingh;i++) {
printf("%c\n", s[i] );
}
return 0;
}
the output I expect is something like s[] = {0,x,f,3}
This computation returns a char consisting of 4.
You are saying your computation returns a character which consists of 4 characters. Is it possible???
I guess your computation returns a string that consists of 4 characters. Example 0xf3
You can not store a string literal in char type variable.
char x = "0xf3"; //<---------------------
You can store them in a character array like char x[] = "0xf3";
Here is the modified code
#include <stdio.h>
#include<string.h> //<---- for strlen() and strcpy()
int main()
{
char x[] = "0xf3"; //<----store in a char array
char s[10];
strcpy(s, x); //<----copy in another array
int lingh = strlen(s);
for (int i = 0; i<lingh;i++) {
printf("%c\n", s[i] );
}
return 0;
}
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * reverse(char *string);
int main(int argc, char *argv[])
{
char array[10];
array[0] = 'a';
array[1] = 'b';
array[2] = 'c';
array[3] = 'd';
array[4] = 'e';
printf("1%s\n",array);
char *p = reverse(array);
printf("4%s\n",p);
printf("5%s\n",array);
}
char * reverse(char *string)
{
int size = strlen(string);
char reversed[size];
int i;
int j = 0;
for(i = size-1; i >= 0; i--)
{
reversed[j] = string[i];
j++;
}
printf("2%s\n",reversed);
string = reversed;
printf("3%s\n",string);
return reversed;
}
This code basically just initializes an array of values and passes it into a method that reverses these values.
I am not sure if this is the best way to execute the task, since I am new to pointers and arrays in C.
But the real question is this:
Can anyone figure out why in this line
printf("4%s\n",p);
if you remove the preceding '4', so it looks like so
printf("%s\n",p);
the line won't print at all?
You are returning a pointer to local variable(reversed) in the function reverse the question should actually be: Why did it work in the first place?.
This code string = reversed; will only copy the pointer, and again the local copy of the pointer so it has no effect outside the function.
To reverse a string you don't need additional memory - this can be done in-place.
Strings in C must end with the null character. You're using strlen on a non null-terminated string.
Furthermore, you just a very lucky person, because there is a serious problem with you code: you forget to add \0 symbol at the end of string.
UPD: the main problem is with code line char reversed[size];.
It's a regular local variable, it has automatic duration, which means that it springs into existence when the function is called and disappears when the function returns (see this link).
You need to change it to:
char *reversed = malloc((size+1)*sizeof(char));
UPD-2: another bug fixing will be:
1) add array[5] = '\0'; after all other array initializing lines
2) add reversed[j] = '\0'; after for...loop:
for(i = size-1; i >= 0; i--)
{
reversed[j] = string[i];
j++;
}
reversed[j] = '\0';
UPD-3: But in general it will much more correctly initialize your string in appropriate way:
char array[10] = "abcde";
This problem is driving me crazy, I'm sure I'm missing something. I need to initialize an array of chars using only pointers. Below is the code I have so far:
int p2(){
/* Implements problem 2 of lab */
// Create an array
char **s = (char**)malloc( 11 *sizeof(char));
char *p = *s;
char start ='A';
while( p != s+10){
*p = start;
start++;
p++;
}
return(0);
}
The problem I'm having is I don't know how to address the characters inside of the array. I understand the base address of the array is **s, and the pointer to the first element is *s. What I don't understand is how to get to **s+10 (i.e. the end of the array).
Can anyone shine some light for me??? Please!
EDIT: Ok, looks like I misunderstood the question. I appears I need to create an array of strings (thus the char ** allocation). Then I need to loop through this array, and assign each string (i.e. char *) a value 15 chars long. Please let me know if I'm understanding this correctly:
char **strings ==> strings[0 ... n ] where each element is a pointer to a char (possibly an array). There for *string ==> strings[0], *(string+1) = strings[1], etc etc.
Am I close or way off?
char **s is 2 dimensional array of characters, or array of C strings if you want.
If you want to use array of characters you should use:
char *string = (char*)malloc( 11 *sizeof(char));
If you really want to initialize array of strings, at first step you're initializing array of pointers, that's:
char **s = (char**)malloc( 11 *sizeof(char *));
Please note that I'm using char * inside sizeof. Than when you may use strings, but at first you must initialize each string.
s[0] = (char*) malloc( 15*size(char)); // This is actually string, 14 characters long (NULL terminated)
char *p = s[0]; // p is pointer to beginning of your string now
And there's two way how to address your string:
s[0][3] // 4th character of your string
p[3]
Or if you want to use just pointers:
char *p = *(s+0);
*(p+3); // 4th character
*((*(s+0))+3) // To do it "hardcore"
EDIT: added an example
When you have **char p and use p++ or p + 1, C increases memory address. *p operator tells compiler that you now want to work with data stored in memory, not with pointer. Therefor those two syntax do the same:
p[10] = 'a';
*(p+10) = 'a';
So if you want traverse both your dimensions, you should use:
for( int i = 0; i < 11; i++){
char *p = *(s+i);
for( int j = 0; j < 10; j++){
*(p + j) = 'a'; // Say you wanna fill them with as
}
// You may also use this syntax:
while( p < (*(s+i) + 10)){ // or use != instead of <
*p = 'a';
p++;
}
}
I think you meant this:
char *s = (char*) malloc(11 *sizeof(char));
char *p = s;
In which case you'd address the characters with s[x]
First, you don't need a char ** unless you need an array of arrays.
Second, you can get to the end of the array with (*s)[10] or *((*s)+10)
Third, in C programming, don't cast the result of malloc()
The code is falling apart here.
char **s = (char**)malloc( 11 *sizeof(char));
You're allocating enough memory for 11 chars, which sounds like what you want to do.
However, you're casting the address to those 11 chars to a (char**) as if you were allocating space for pointers, not chars.
Why should you use a double pointer for creating an array of chars?
char *s = (char *) malloc(10 * sizeof(char));
char *start = s; // save starting pointer
for(; s < start+10; s++)
*s = 'a';
return 0;
If you allocate char ** essentially you're allocating an array of pointers to char(eg. array of strings). If you need char array, allocate char array. If you start working with pointers a design of stack and heap can be really helpful.