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]);
}
Related
#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
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);
}
im very new to programming, trying to learn C and cant figure out how to create/use a simple function.
Im trying to create a function called stringtest and then call it into the main and simply make the string strA print ABC.
void stringtest(char strA[20])
{
strA = "ABC";
}
int main()
{
char strA;
stringtest(strA[20]);
printf("This is strA", strA);
return 0;
}
You need to read up on pointers and the C syntax in general.
This is one way you could do it.
#include <stdio.h>
#include <string.h>
void stringtest(char *strA) {
strcpy(strA, "ABC");
}
int main(int argc, const char * argv[]) {
char strA[20];
stringtest(&strA[0]);
printf("This is strA -> %s \n", strA);
return 0;
}
Take care,
/Anders.
I don't think your code ran!!
There are a lot of bugs and errors in your code.
See the code given below to understand how to do this:
#include <stdio.h>
char strA[20];
void stringtest(){
strA[0]='A';
strA[1]='B';
strA[2]='C';
strA[3]='\0';
}
int main(){
stringtest();
printf("This is strA %s",strA);
}
Case 1:
When I take string input, it successfully gives the output, writing this piece of code:
#include <stdio.h>
int main()
{
char *str;
scanf("%s",&str);
printf("%s",&str);
return 0;
}
Case 2:
On the other hand, it throws a Runtime Error for this snippet:
#include <stdio.h>
int main()
{
char *str;
scanf("%s",&str);
printf("%s",str);
return 0;
}
I found this thing peculiar, and want to know why it happens...
Thanks in advance.
None of those two cases are right.
Case 1 only worked because you got lucky, probably by giving a short string as input. Try something like "bfjabfabjkbfjkasjkvasjkvjksbkjafbskjbfakbsjfbjasbfjasbfkjabsjfkbaksbfjasbfkja" and you'll suffer a seg fault, most likely.
You should have a block of memory associated with str, either on the stack by declaring an array for it or on the heap malloc'ing memory for it.
And you shouldn't use the & operator.
So it would go like this:
#include <stdio.h>
int main()
{
char str[50]; // 50 is arbitrary
scanf("%s",str);
printf("%s",str);
return 0;
}
or like this:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char* str = malloc(50); // 50 is arbitrary
scanf("%s",str);
printf("%s",str);
free(str);
return 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;
}