I am starting to learn about pointers in C.
How can I fix the error that I have in function x()?
This is the error:
Error: a value of type "char" cannot be assigned to an entity of type "char *".
This is the full source:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdarg.h>
void x(char **d, char s[]) {
d = s[0]; // here i have the problem
}
void main() {
char *p = NULL;
x(&p, "abc");
}
In function x() you pass d which is a char ** (a pointer to a string pointer), and char s[] (an array of char, passed as similarly to a pointer to char).
So in the line:
d = s[0];
s[0] is a char, whereas char **d is a pointer to a pointer to char. These are different, and the compiler is saying you cannot assign from one to the other.
However, did your compiler really warn you as follows?
Error: a value of type "char" cannot be assigned to an entity of type "char *"
Given the code sample, it should have said char ** at the end.
I think what you are trying to make x do is copy the address of the string passed as the second argument into the first pointer. That would be:
void x(char **d, char *s)
{
*d = s;
}
That makes p in the caller point to the constant xyz string but does not copy the contents.
If the idea was to copy the string's contents:
void x(char **d, char *s)
{
*d = strdup(s);
}
and ensure you remember to free() the return value in main(), as well as adding #include <string.h> at the top.
A more proper way would be to use strcpy:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdarg.h>
void x(char **d) {
*d = malloc(4 * sizeof(char));
strcpy(*d, "abc");
}
int main() {
char *p;
x(&p);
printf("%s", p);
free(p);
return 0;
}
Outputs: abc
Related
Basically at the most basic level I cant understand why can't I do this:
#include <stdio.h>
#include <stdlib.h>
void mal(char *str){
str = malloc(sizeof(char));
}
int main(void)
{
char *s;
mal(s);
free(s);
return 0;
}
s gets passed by value to the function mal. Inside mal the local parameter str gets changed by the assignment, but s in main is kept uninitialized. In C you should pass a pointer to s to mal to resolve this problem:
#include <stdio.h>
#include <stdlib.h>
void mal(char **str){ // pointer to pointer
*str = malloc(sizeof(char)); // referenced variable behind str changed
}
int main(void)
{
char *s;
mal(&s); // pointer to s passed
free(s);
return 0;
}
I have this struct:
typedef struct cmdLine {
char * const arguments[256];
} cmdLine;
I also have an argument cmdLine *pCmdLine. I want to use execvso I write execv((pCmdLine->arguments[0]), pCmdLine->arguments);. The second argument doesn't feet properly to execvand I want to ask how to convert it properly.
The warning I get is: Passing 'char* const[256]'' to parameter of type 'const char *const *' discards qualifiers in nested pointer types. I would lie for some help to convert it properly, thanks.
I do not see any problems:
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <stdio.h>
struct
{
char * const arr[20];
}*str;
void foo(char *const par[])
{
volatile const char * const ptr = par[4];
printf("%s\n", par[7]);
printf("%s\n", ptr);
}
void foo1()
{
foo(str -> arr);
}
https://godbolt.org/z/4Sv4a8
I'm new to C and I'm having a hard time understanding how to call methods with pointers. Currently this code should reverse a null-terminated string, but I get the errors
main.c:8:12: error: use of undeclared identifier 'sas'
char* N = sas;
^ main.c:10:10: warning: incompatible integer to pointer conversion passing 'char' to parameter of type
'char *'; remove * [-Wint-conversion]
reverse(*N);
^~ ./header.h:3:27: note: passing argument to parameter 'N' here EXTERN void reverse(char *N);
My actual code is this:
Main:
#include <stdio.h>
#include <stdlib.h>
#include "header.h"
int main(int argc, char *argv[])
{
char* N = sas;
reverse(*N);
}
reverse:
#include <stdio.h>
#include "header.h
#include <stdlib.h>
void reverse(char *str)
{
char* end = str;
char temp;
printf("this is *str: %c\n", *str);
if (str)
{
while (*end)
{
++end:
}
end--;
while (str < end)
{
temp = *str
*str++ = *end;
*end-- = temp;
}
}
}
header.h:
#define EXTERN extern
EXTERN void reverse(char *N)
thanks for the help and time!
int main(int argc, char *argv[])
{
char* N = "sas";
reverse(*N);
}
First you make N point to a string constant. Then you try to reverse what N points to. But since N points to a string constant, you're trying to reverse a string constant. By definition, constants cannot have their values changed.
First of all, there're many syntax errors within that piece of code:
'sas' what is that? your compiler thinks it's a variable but can't find any with that name. if you wanted to put a "sas" string, then:
char* N = "sas";
inconsistant brackets. More closing brackets than opening ones, and no opening bracket after declaring your function.
As I understand it, you are trying to reverse a string. This is just a slight modification of your code.
Full Source:
#include <stdio.h>
#include <stdlib.h>
void reverse(char *N);
int main(int argc, char *argv[])
{
char strSas[] = "sas";
reverse(strSas);
}
void reverse(char *str)
{
char* end = str;
char temp;
if(str) {
printf("this is *str: %c\n", *str);
while(*end) {
++end;
}
end--;
while(str < end) {
temp = *str;
*str = *end;
*end = temp;
++str;
--end;
}
}
}
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;
}