#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <regex.h>
#include <unistd.h>
#include <ctype.h>
#include <assert.h>
void *process(char **nbE)
{
char buffer[8] = "test";
*nbE = &buffer[0];
printf("%s\n", *nbE);
}
int main(int argc, char **argv)
{
char *str;
process(&str);
printf("%s\n", str);
}
I'm trying to get the value of *nbE in main() by making it points to the address of first char in my array.
But it returns something not encoded, why?
What would be a way for me to do this way?
Note: I know I can do it simpler, I have a more complex code and this is a mini example
Basically I have something interesting in my array and want to pass it to my main function through a char* variable
char buffer[8] = "test";
creates a string that is local to the function, it is destroyed once you return from that function. Do this
static char buffer[8] = "test";
or
char * buffer = strdup("test");
you have to release the string when you have finsihed with it in the second case
Related
i stucked, need to add pointer char value to list char value. pointer char value includes username (windows machine) and list char value includes startup path. I tried to strcat but it breaks the first ";" part.
#define _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
char *username = getenv("USERNAME");
//printf("%s\n", username);
char shortcutpath[100] = "C:\\Users\\";"\\AppData\\Roaming\\Microsoft\\Windows\\Start Menu\\Programs\\Startup"; //2 ; is weird, i just tried
strcat(shortcutpath, username);
printf("%s\n",shortcutpath);
return 0;
}
Output = C:\Users%username% ,needs to continue
Need to assign a variable "C:\Users\%username%\AppData\Roaming\Microsoft\Windows\Start Menu\Programs\Startup" this path
Is there any easy way to do that? Or something other usefull things?
You probably want this:
#define _CRT_SECURE_NO_WARNINGS
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
char* username = getenv("USERNAME");
char shortcutpath[100] = "C:\\Users\\";
strcat(shortcutpath, username);
strcat(shortcutpath, "\\AppData\\Roaming\\Microsoft\\Windows\\Start Menu\\Programs\\Startup");
printf("%s\n", shortcutpath);
return 0;
}
But the more correct way would probably be to get directly the APPDATA environnement variable:
int main(void)
{
char shortcutpath[100];
char* appdata = getenv("APPDATA"); // C:\Users\<yourusername>\AppData\Roaming
strcpy(shortcutpath, appdata);
strcat(shortcutpath, "\\Microsoft\\Windows\\Start Menu\\Programs\\Startup");
...
char shortcutpath[255] = "C:\\Users\\";
strcat(shortcutpath, username);
strcat(shortcutpath, "\\AppData\\Roaming\\Microsoft\\Windows\\Start Menu\\Programs\\Startup");
This code compiles without errors, but upon opening the app, it says:
file.exe has stopped working
#include <string.h>
#include <stdio.h>
int main() {
char *a = 'Hello';
char *b = 'World';
strcat(a,b);
puts(a);
}
Where did I go wrong?
You need to allocate sufficient space and use double quote instead of single quote. You could use array.
#include <string.h>
#include <stdio.h>
int main() {
char a[20] = "Hello";
char b[10] = "World";
strcat(a,b);
puts(a);
}
Constant strings are not modifiable. This is a proper way to declare, initialize and modify a string buffer in C:
#include <string.h>
#include <stdio.h>
int main() {
char a[20];
char *b = "World";
strcpy(a,"Hello");
strcat(a,b);
puts(a);
return(0);
}
You can't do strcat on pointer of characters. You only can do strcat on array of characters .... Sorry for my precedent answer, look at the code below :
#include <string.h>
#include <stdio.h>
int main() {
char a [20];
char b[20];
strcpy(a,"Hello");
strcpy(b,"World");
strcat(a,b);
puts(a);
return(0);
}
I am trying to add an int to a Multi-dimensional char Array. After reading the link below I would think I can use sprintf. If I can't use sprintf what is another way I can do this?
http://www.cplusplus.com/reference/cstdio/sprintf/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int main(int argc, char *argv[])
{
//{"TYPE", "ID", "SCOPE", "VALUE"}
char *symbol_table_variables[503][4];
int scope = 0;
int lower_bound_of_big_boy_counter = 0;
sprintf (symbol_table_variables[lower_bound_of_big_boy_counter][2], "%d", scope);
printf("symbol_table_variables[lower_bound_of_big_boy_counter][2] %s \n",
symbol_table_variables[lower_bound_of_big_boy_counter][2]);
return 0;
}
An update.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
int main(int argc, char *argv[])
{
//{"TYPE", "ID", "SCOPE", "VALUE"}
char *symbol_table_variables[503][4] = {0};
int scope = 5;
int lower_bound_of_big_boy_counter = 0;
char scope_char[80] = {0};
sprintf (scope_char, "%d", scope);
printf("scope_char %s \n", scope_char);
symbol_table_variables[lower_bound_of_big_boy_counter][2] =
malloc(strlen(scope_char)+1);
strcpy(symbol_table_variables[lower_bound_of_big_boy_counter][2],
scope_char);
memset(scope_char, 0, 80);
//sprintf (symbol_table_variables[lower_bound_of_big_boy_counter][2], "%d", scope);
printf("symbol_table_variables[lower_bound_of_big_boy_counter][2] is %s \n",
symbol_table_variables[lower_bound_of_big_boy_counter][2]);
return 0;
}
symbol_table_variables[lower_bound_of_big_boy_counter][2] has no memory allocated to it you you are invoking undefined behavior.
One solution would be to allocate some memory
symbol_table_variables[lower_bound_of_big_boy_counter][2] = malloc(32);
printf (symbol_table_variables[lower_bound_of_big_boy_counter][2], "%d", scope);
That isn't great because you don't really know how much memory you need.
I'd be questioning the need for a 2D array of strings...
I need to put 3 strings on an array[3][3].
I tried to do it with pointers, but I only receive a single character.
#include <stdio.h>
int array[3][3]
char thing[5] = "thing";
main()
{
thing = array[0][0];
printf("%s", array[0][0];
}
Try this. With due respect your code absolutely incorrect and need many changes. You need to update your programming skills too.
#include <stdio.h>
#include <string.h>
char array[3][6]={0};
char *thing = "this";
main()
{
strcpy(array[0],thing);
printf("%s\n", array[0]);
}
This question already has answers here:
Passing address of array as a function parameter
(6 answers)
Closed 9 years ago.
I'm writing a function that gets a string, allocates memory on the heap that's enough to create a copy, creates a copy and returns the address of the beginning of the new copy.
In main I would like to be able to print the new copy and afterwards use free() to free the memory. I think the actual function works although I am not the char pointer has to be static, or does it?
The code in main does not work fine...
#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
int make_copy(char arr[]);
int main()
{
char arrr[]={'a','b','c','d','e','f','\0'};
char *ptr;
ptr=make_copy(arrr);
printf("%s",ptr);
getchar();
return 0;
}
int make_copy(char arr[])
{
static char *str_ptr;
str_ptr=(char*)malloc(sizeof(arr));
int i=0;
for(;i<sizeof str_ptr/sizeof(char);i++)
str_ptr[i]=arr[i];
return (int)str_ptr;
}
OK, so based on the comments. A revised version:
#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
char* make_copy(char arr[]);
int main()
{
char arrr[]={"abcdef\0"};
char *ptr=make_copy(arrr);
printf("%s",ptr);
getchar();
return 0;
}
char* make_copy(char arr[])
{
static char *str_ptr;
str_ptr=(char*)malloc(strlen(arr)+1);
int i=0;
for(;i<strlen(arr)+1;i++)
str_ptr[i]=arr[i];
return str_ptr;
}
Or even better:
#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
char* make_copy(char arr[]);
int main()
{
char arrr[]={"abcdef\0"};
printf("%s",make_copy(arrr));
getchar();
return 0;
}
char* make_copy(char arr[])
{
char *str_ptr;
str_ptr=(char*)malloc(strlen(arr)+1);
return strcpy(str_ptr,arr);
}
You're on the right track, but there are some issues with your code:
Don't use int when you mean char *. That's just wrong.
Don't list characters when defining a string, write char arrr[] = "abcdef";
Don't scale string alloations by sizeof (char); that's always 1 so it's pointless.
Don't re-implement strcpy() to copy a string.
Don't cast the return value of malloc() in C.
Don't make local variables static for no reason.
Don't use sizeof on an array passed to a function; it doesn't work. You must use strlen().
Don't omit including space for the string terminator, you must add 1 to the length of the string.
UPDATE Your third attempt is getting closer. :) Here's how I would write it:
char * make_copy(const char *s)
{
if(s != NULL)
{
const size_t size = strlen(s) + 1;
char *d = malloc(size);
if(d != NULL)
strcpy(d, s);
return d;
}
return NULL;
}
This gracefully handles a NULL argument, and checks that the memory allocation succeeded before using the memory.
First, don't use sizeof to determine the size of your string in make_copy, use strlen.
Second, why are you converting a pointer (char*) to an integer? A char* is already a pointer (a memory address), as you can see if you do printf("address: %x\n", ptr);.
sizeof(arr) will not give the exact size. pass the length of array to the function if you want to compute array size.
When pass the array to function it will decay to pointer, we cannot find the array size using pointer.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char *strdup(const char *str)
{
char *s = (char*)malloc(strlen(str)+1);
if (s == NULL) return NULL;
return strcpy(s, str);
}
int main()
{
char *s = strdup("hello world");
puts(s);
free(s);
}
Points
~ return char* inside of int.
~ you can free the memory using below line
if(make_copy!=NULL)
free(make_copy)
Below is the modified code.
#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
char* make_copy(char arr[]);
int main()
{
char arrr[]={'a','b','c','d','e','f','\0'};
char *ptr;
ptr=make_copy(arrr,sizeof(arrr)/sizeof(char));
printf("%s",ptr);
printf("%p\n %p",ptr,arrr);
getchar();
return 0;
}
char* make_copy(char arr[],int size)
{
char *str_ptr=NULL;
str_ptr=(char*)malloc(size+1);
int i=0;
for(;i<size;i++)
str_ptr[i]=arr[i];
str_ptr[i]=0;
return str_ptr;
}