Its a simple input that I want to make for char*. Why is this not working?
It throws me an exception that I can't resolve..
#include "stdafx.h"
#include "stdio.h"
#include "stdlib.h"
char* GetCharSeq()
{
char *s = (char*)malloc(100);
scanf_s("%s", s);
return s;
}
int main()
{
char* charseq;
charseq = GetCharSeq();
return 0;
}
You have undefined behavior in your code. You have it because you provide to few arguments to the scanf_s function.
For every string argument, you need to provide not only the destination string but also the size of the string. So change your call to
scanf_s("%s", s, 100);
modify your code
char* GetCharSeq()
{
char *s = (char*)malloc(100);
gets(s);
return s;
}
This will work.
Related
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;
}
Is there a way to make a function that stores a string, and then call that function in int main() that displays it on the screen? I have been searching a lot and haven't found a clear example. Here is my code. I would like to be able to call it without using the if statement
#include <stdio.h>
/* function declaration */
int StrPrint(char *str);
/* main() function */
int main()
{
char str[] = "The string i am returning n";
int (*ptr)(char *str);
ptr = StrPrint;
if (!(*ptr)(str))
printf("Done!\n");
return 0;
}
/* function definition */
int StrPrint(char *str)
{
printf("%s\n", str);
return 0;
}
The code you've posted is far more complicated than the simple task you're trying to accomplish.
Why not something like this:
#include <stdio.h>
void StrPrint(char* str);
int main(void)
{
char str[] = "The string i am returning n";
StrPrint(str);
return 0;
}
void StrPrint(char* str)
{
printf("%s\n", str);
}
This conflicts slightly with your requirement, in that the function doesn't store a string it just prints out the string passed to it as an argument. But according to the code you posted, this looks like what you're trying to accomplish.
As mentioned by you if you don't like to use pointers it can be done as shown below:
#include <stdio.h>
int StrPrint(char s[])
{
printf("%s\n", s);
return 0;
}
int main()
{
char str[] = "The string i am returning n";
if (!StrPrint(str))
printf("Done!\n");
return 0;
}
Here's the code:
#include <stdio.h>
#include <string.h>
void print (void*);
int main (void)
{
char *a = "Mcwhat";
print(&a);
printf("\n%s", a);
return 0;
}
void print (void *text)
{
char* pchar[5];
*pchar = (char*)text;
strcpy( *pchar, "Mcthat" );
}
I am trying to make Mcwhat into Mcthat using a void parameter, but the printf gives me a segmentation fault afterwards. Where is my mistake? I managed to do it char by char but now I want to change the whole string. Didn't found enough material on this in the books on C I have.
Keep it simple and pay attention to the type of your variables :
#include <stdio.h>
#include <string.h>
void print (void*);
int main()
{
char a[] = "Mcwhat"; // a is now a read-write array
print(a); // a decays to a pointer, don't take its adress or you'll get a pointer-to-pointer
printf("\n%s", a);
return 0;
}
void print (void *text)
{
strcpy( text, "Mcthat" ); // Don't dereference text here
}
Note that this "print" function is unsafe in all imaginable ways, but that wasn't the question.
There are lot of issues in your code:
1. Char array should be big enough to store the string. char[5] cannot hold Mswhat.
2. char* pchar [5] declares 5 char pointers, whereas you need one char pointer pointing to a char array.
I have corrected it.
#include <stdio.h>
#include <string.h>
void print (char*);
int main (void)
{
char *a = malloc(10);
strcpy(a,"Mcwhat");
print(a);
printf("\n%s", a);
free(a);
return 0;
}
void print (char *text)
{
char *pchar = text;
strcpy( pchar, "Mcthat" );
}
Just write it like that
void print (char *text)
{
strcpy( text, "Mcthat" );
}
But make sure, the that size of text is large enough to put "Mcthat" inside it.
Also in main:
print(a);
instead of
print(&a); // would requite void print (char** text)
tho whole shebang:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void print (void*);
int main (void)
{
char *a = malloc(strlen("Mcwhat")+1);
print(a);
printf("\n%s\n", a);
free(a);
return 0;
}
void print (void *text)
{
strcpy(text, "Mcthat" );
}
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;
}