I have a source pointer (pSource) and a goal pointer (pGoal). I also have a number of characters (n) that need to be copied to the pGoal from pSource.
I thought that I can just copy what's in the pSource to pGoal and move both pointers to the next location. (Both are pointing at the start at the beginning).
for (int i = 0; i < n; i++) {
pGoal+i = pSource+i;
}
Assuming that your pointers are of type char *, the correct way of doing this is:
for (int i = 0; i < n; i++) {
*(pGoal+i) = *(pSource+i);
// or pGoal[i] = pSource[i]
}
You can also check memcpy
Related
I have to arrays of int for example arr1={0,1,1,0,0}, arr2={1,0,1,1,1} and I need to return 1 char* created by malloc that will be shown like this : "01100,10111".
when I do for loop it doesn't work, how can I do it ?
char* ans = (char*)malloc((size * 2+1) * sizeof(int));
for (int i = 0; i < size; i++)
ans[i] = first[i];
ans[size] = ",";
for (int i = size+1; i < 2*size+1; i++)
ans[i] = second[i];
Among the multitude of problems:
Your allocation size is wrong. It should include space for the separating comma and the terminating nullchar. sizeof(int) is wrong regardless, it should be sizeof(char) and as-such can be omitted (sizeof(char) is always 1).
Your storage is wrong. You want to store characters, and your values should be adjusted relative to '0'.
Your indexing of the second loop is wrong.
In reality, you don't need the second loop in the first place:
char* ans = malloc(size * 2 + 2);
for (int i = 0; i < size; i++)
{
ans[i] = '0' + first[i];
ans[size+1+i] = '0' + second[i];
}
ans[size] = ',';
ans[2*size+1] = 0;
That's it.
1.
char* ans = (char*)malloc((size * 2+1) * sizeof(int));
What is size here? It is not defined and declared in the provided code.
You do not need to cast the return value of malloc() to char. In fact, you do not need to cast the return value of malloc() anymore. It is a habit from the early C days.
Why do you need a char pointer here at all exactly? If you want to print 01100,10111 there is no need to use a char pointer for the output of the integer values.
2.
for (int i = 0; i < size; i++)
ans[i] = first[i];
Again what is size here?
What is first here? If it isn´t a pointer this statement is invalid.
3.
ans[size] = ",";
This operation is invalid. You are trying to assign a string to a pointer.
By the way, I don´t know what you trying to do with this statement. You can incorporate the comma separate in the output of 01100,10111, without your intend to include it int the memory of the int arrays itself.
4.
for (int i = size+1; i < 2*size+1; i++)
ans[i] = second[i];
Same as above: What is value and the type of size?
What is second? If it isn´t it a pointer this statement is invalid.
5.
To answer to the question title:
(How to) Copy two arrays of int to one char* in C
This isn´t possible. You can´t copy two arrays with its data to a pointer to char.
There are at least four issues with your code.
You malloc the wrong size, you want to use sizeof(char).
You need to zero terminate it, so you need to add extra room for the terminating zero
char* ans = (char*)malloc((size * 2+2) * sizeof(char));
second[size * 2+1] = 0;
Also the indexing of the second loop is wrong. You are accessing second array out of bounds. Make the loop more like the first.
We also need to convert the integer value to a char in the loops.
for (int i = 0; i < size; i++)
ans[size+i+1] = second[i] + '0';
I'm Trying to check in my matrix of dimension [10][10], which spots are available to store data (String) there and which are occupied.
The code basically goes through the whole matrix and checks every spot.
I have tried using the strlen and != NULL but everything just prints that the spot is free.
char parque[10][10];
for(int i = 0; i < 10; i++) {
for(int j = 0; j < 10; j++) {
parque[i][j] = "";
}
}
parque[5][5]="f47ac10b-58cb-4372-a567-0e02b2c3d499,ANR";
for(int i = 0; i < 10; i++) {
for(int j = 0; j < 10; j++) {
if(parque[i][j] != "") {
printf("The Spot [%d][%d] is taken",i,j);
} else {
printf("The Spot [%d][%d] is free",i,j);
}
}
}
Basically the spot [5][5] should print that it's taken, at least that's what I want it to do...
Thanks in advance!
Your declaration
char parque[10][10];
declares a two-dimensional array of char. If you compile your code with a strict compiler, you'll get an error:
error: assignment makes integer from pointer without a cast [-Wint-conversion]
parque[i][j] = "";
^
What you did mean is to make an array of pointers to const char, like here:
const char* parque[10][10];
Then your program will say that The Spot [5][5] is taken.
You can't use !=. You need to use strcmp. And, of course, you need to initialize your array content before iterating it and using its values to compare with "" string.
This condition:
if(parque[i][j] != "")
Will become:
if (strcmp(parque[i][j], ""))
void memSet(char destination[], char valueMemSet, int numOfValue)
{
char temp;
int j=1;
for (int i = 0; i <= numOfValue; i++)
{
temp = destination[i];
destination[i] = valueMemSet;
destination[j] = temp;
j++;
}
}
The array is originally "this is the source Concatenate means to link."
This is what I am trying to get "------this is the source Concatenate means to link."
This is What I am currently getting "-------Tthe source Concatenate means to link."
When I ran the debugger it saves the first letter of the array but every single one after gets replaced.
How can I solve this issue?
memset() is a function, which sets a particular value in a given memory, like you want to initialize total array's elements to some particular value(for eg - zero). So it will set the same in that array.
what you need here is strcat() function.
Is that what you need?
void memSet(char destination[], char valueMemSet, int numOfValue, int len)
{
int j=len-numOfValue;
for (int i = len-1;i>=numOfValue;i--) {
destinstion[i] = destination[j--];
}
for (int i = 0; i < numOfValue; i++)
{
destination[i] = valueMemSet;
}
}
I am trying to find length of pointer array.
Here is my code:
char **histargv[histsize];
histargv[hist] = (char**)malloc(sizeof(char *)*arg);
variable arg is not constant. Each index of histargv array has different arg value.
So how can I find arg value for each index?
Thanks
If the pointers are for strings, then you can figure out the length of the strings (not the size of the memory block) by using strlen. I.e. with:
for (size_t i = 0; i < 100; ++i) {
histargv[i] = read_some_string_somehow_and_return_in_new_buffer();
}
you can print the string including their lengths like this:
for (size_t i = 0; i < 100; ++i) {
printf("%3zu %zu %s\n", i, strlen(histargv[i]), histargv[i]);
}
If this are not strings, then you should follow the link in Eli Sadoffs comment.
Regards
I want to remove some elements from my array and re-size it.
for example my array is:
char get_res[6] = {0x32,0x32,0x34,0x16,0x00,0x00};
Now I want to remove elements after 0x16, so my desire array is:
get_res[] = {0x32,0x32,0x34,0x16};
what is solution?
You cannot resize arrays in C (unlike Python, for example). For real resizing, at least from an API user's point of view, use malloc, calloc, realloc, and free (realloc specifically).
Anyway, "resizing" an array can be imitated using
a delimiter; for example, a delimiter like 0xff could mark the end of the valid data in the array
Example:
#define DELIMITER 0xff
print_data(char* data) {
for (size_t i = 0; data[i] != DELIMITER; ++i)
printf("%x", data[i]);
}
a member counter; count the number of valid data from the beginning of the array onward
Example:
size_t counter = 5;
print_data(char* data) {
for (size_t i = 0; i < counter; ++i)
printf("%x", data[i]);
}
Notes:
Use unsigned char for binary data. char may be aliasing signed char, which you might run into problems with because signed char contains a sign bit.
There is no need to "remove" them. Just don't access them. Pretend like they don't exist. Same like in stacks, when you "pop" a value from the top of the stack, you just decrement the stack pointer.
Manipulating arrays in C isn't easy as it is for vector in C++ or List in Java. There is no "remove element" in C. I mean that you have to do the job yourself, that is, create another array, copy only the elements you want to this new array, and free the memory occupied by the previous one.
Can you do that? Do you want the code?
EDIT:
Try that. It's just a simple program that simulates the situation. Now, you have to see the example and adapt it to your code.
#include <stdio.h>
#include <stdlib.h>
int main() {
char get_res[6] = {0x32,0x32,0x34,0x16,0x00,0x00};
char target = 0x16;
int pos, i, length = 6; // or specify some way to get this number
for(i = 0; i < length; i++)
if(get_res[i] == target) {
pos = i;
break;
}
pos = pos + 1; // as you have to ignore the target itself
char *new_arr = malloc(pos);
for(i = 0; i < length; i++) {
new_arr[i] = get_res[i];
i++;
}
for(i = 0; i < pos; i++)
printf("%c ", new_arr[i]);
return 0;
}