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I have a little problem with casting int to char* (string)... is it even possible in C?
I'll try to explain why i need this.
I can cast int to char but I need cast int to char*.
I had a int varriable (int number_of_revisions)
and I need convert this number of revisions to char * becouse I need create a name of file and the number of revision is part of the name.... so there is part of code for better imagination of this problem.
int number_of_revision = 970; // 970 just for example
char * version;
char * new_name;
char ch_number_of_rev[4];
version = "0.";
itoa(number_of_revision,ch_number_of_rev,10);
//strcat(version, ch_num_o_rev ); // doesn't work becouse ch_number_of_rev is char and strcat requires char*
please I need quick help... Have anybody any idea how to do it? ...
but I need cast int to char*
Casting only changes the type - it does not change the value within the variable. If you need to convert an int to array of chars (i.e. a string) then use sprintf or snprintf:
char* buffer = ... allocate a buffer ...
int value = 970;
sprintf(buffer, "%d", value);
Converting int to string in c
Also, you have not allocated any memory for version - use malloc and allocate some memory.
strcat here won't work because you haven't allocated any space to store the result in. Your version is probably in read-only memory anyway, so you'd get a segfault, otherwise you'll get memory corruption. So make sure to allocate enough space for it, e.g. by using
char version[10] = "0.";
You may want to read up on pointers first, though.
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Old Test
I'm going over a test I took and am trying to figure out what the answer was to these questions. I was wondering if anyone could help me? As you can probably see I did not really understand how to answer them at the time but I would like to learn. I believed the answer has something to do with Malloc, but was unsure exactly how.
Thank you!
Edit : Is this how you do it?
#include <stdio.h>
#include <stdlib.h>
float* func();
int main(void)
{
float *x;
x = func();
printf("%f\n", *x);
return 0;
}
float* func(void){
float * z;
z = malloc(sizeof(float));
* z = 11.2;
return z;
}
malloc is related to allocating memory.
When we talk about array and pointer in c, we can seperate it into static array and dynamic array. For static array, we use array, for example,
char arr[10];
which means declare char type array named arr with length of 10.
For Dynamic array, we use pointer, for example, char *arr. This means char type pointer of arr. Pointer is very flexible; therefore, you must command to use it properly.
Assume
char *arr = (char *) malloc (sizeof (char) * 10);
This means you have a pointer and will allocate the memory with the size of char type with length of 10 you can also re allocate memory with realloc with different length. At the end of using it you must
free(arr);
To add, this is benefit of C language and I believe it is harder to use than other languages but more flexibility. On the other hand, you must be very very careful using it. Unproperly used pointer could cause entire software failure.
As float z is defined locally in fucntion, it's allocated on stack.
As a result it memory allocation is destroyed when function exits.
As a result you will have a runtime error cause you are accesing a memory that do not belongs to you.
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i need to detect the problem in the next code, and the reason to that problem and how to fix it. for some reason when i tried to run it in visual the error is on the free.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <malloc.h>
int main()
{
char str1[] = "abcde";
char* str2 = (char*)malloc(strlen(str1));
strcpy(str2, str1);
puts(str1);
puts(str2);
free(str2);
return 0;
}
strlen return the length of null terminated string excluding the null character '\0'. You need to allocate space for null character too.
char* str2 = malloc(strlen(str1) + 1); // Do not cast return value of malloc
it should be
char* str2 = (char*)malloc(strlen(str1)+1);
The issue, as others have mentioned, is that you're not allocating enough space for the string you want to copy. strlen returns the number of characters in the string, however that number doesn't include the null byte at the end that terminates the string.
So when you call strcpy, you're writing one byte past the end of the allocated memory. Once you write past your memory bounds, that invokes undefined behavior. That means your program might appear to work, it might crash (sometimes right away, sometimes later), or it might cause data corruption that would be hard to detect.
In this particular situation, the extra byte you wrote probably corrupted data used by the implementation of free and/or malloc. But with some other compiler or OS, it might work fine.
So to avoid undefined behavior, be sure to allocate the required amount of space:
char* str2 = malloc(strlen(str1) + 1);
Also, don't cast the return value of malloc, as that may mask other errors in your code.
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it seems that this is a duplicate question, but I searched stackoverflow's question about that point and non is like to my problem(I think)
I've two variables of a struct each has its own pointer to char, when I tried to copy from one variable's string to another variable's string, nothing happened, although no errors appear, just warning
implicit declaration of function strcpy
incompatible implicit declaration of built-in function 'strcpy'
I read from some questions on stackoverflow that you'd better to use strdup() function instead of strcpy() but when I did, I had an error
too many arguments to function 'strdup'
I read that there's a problem with strcpy() called "segmentation fault" and I knew it's about memory allocation, I don't totally understand what it's exactly and don't know if it's the problem with my code?
and this is my code
struct p
{
char *name;
struct p *next;
};
struct p *ptrary[10];
int main(void)
{
struct p p,q;
p.name="xyz";
p.next=NULL;
ptrary[0]=&p;
strdup(q.name,p.name);
ptrary[1]=&q;
printf("%s\n",ptrary[1]->name);
return 0;
}
so what is the problem and how I can solve it?
strdup() takes only one argument; it malloc's and returns a new block of heap memory containing the duplicated string. (See https://stackoverflow.com/questions/252782/strdup-what-does-it-do-in-c)
Which probably also points to the problem you were having before -- were you remembering to malloc the space for q to copy p's contents into?
change
strdup(q.name,p.name);
to
q.name = strdup(p.name);
see man page of strdup for further details. The strdup() function returns a pointer to a new string.
full code:
#include <stdio.h>
#include <string.h>
struct p
{
char *name;
struct p *next;
};
struct p *ptrary[10];
int main(void)
{
struct p p,q;
p.name="xyz";
p.next=NULL;
ptrary[0]=&p;
q.name = strdup(p.name);
ptrary[1]=&q;
printf("%s\n",ptrary[1]->name);
return 0;
}
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I have a problem in displaying the size of the array correctly.
I know array size is 256000 but it is displaying as 8 when I enter the loop. size will be displayed accurately if dynamic allocation is not used. How do I rectify the bug using dynamic allocation?
This will give you size 10, because the compiler knows it's an array;
char foo[10];
int size = sizeof foo;
This will give you size 4 on a 32-bit architecture, because it's the size of a pointer.
char *foo = malloc(10 * sizeof(char));
int size = sizeof foo;
After this, the usage of foo is identical. You can do foo[2] or *foo or whatever with both versions. But you probably shouldn't take the address of &foo with the 1st variant. And you should free(foo); sometimes with the 2nd.
Always remember: sizeof is not a function, sizeof is always decided in compile time.
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I'm not sure what the following does and i'm hoping someone can clarify the purpose of having the asterisk in front of the functions name:
char *Foo(char *ptr) {
return NULL;
}
I understand that you can pass by value the memory location of something in the function argument call and *ptr would be the pointer to it. I understand you can create a pointer function that can be used to point to other functions like a regular pointer points to variable memory location but in this case this is not a function pointer that we can point to other functions, or is it? This seems like a real function.
Foo is a function.
It has input: ptr of type char*
It has output of type char*
char* means "pointer to char"
it returns NULL.
That is the most plain explanation I can think of.
its misleading you, the * by the name isn't related to the name
it means the same as char* Foo(char* ptr)
which means a function which takes a char* and returns a char*