Can somebody tell me why this program does not work?
int main()
{
char *num = 'h';
printf("%c", num);
return 0;
}
The error I get is:
1>c:\users\\documents\visual studio 2010\projects\sssdsdsds\sssdsdsds\sssdsdsds.cpp(4): error C2440: 'initializing' : cannot convert from 'char' to 'char *'
But if I write the code like that:
int main()
{
char num = 'h';
printf("%c", num);
return 0;
}
it's working.
char *num = 'h';
Here, the letter 'h' is a char, which you are trying to assign to a char*. The two types are not the same, so you get the problem that you see above.
This would work:
char *num = "h";
The difference is that here you're using double-quotes ("), which creates a char*.
This would also work:
char letter = 'h';
char* ptrToLetter = &letter;
You should read up on pointers in C to understand exactly what these different constructions do.
char * is a pointer to a char, not the same thing than a single char.
If you have char *, then you must initialize it with ", not with '.
And also, for the formatting representation in printf():
the %s is used for char *
the %c is only for char.
In thefirst case you declared num as a pointer to a char. In the second case, you declare it as a char. In each case, you assign a char to the variable. You can't assign a char to a pointer to a char, hence the error.
'h' = Char
"h" = Null terminated String
int main()
{
char *num = "h";
printf("%s", num); // <= here change c to s if you want to print out string
return 0;
}
this will work
As somebody just said, when you write
char *num = 'h'
The compiler gives you an error because you're trying to give to a pointer a value. Pointers, you know, are just variables that store only the memory address of another variable you defined before. However, you can access to a variable's memory address with the operator:
&
And a variable's pointer should be coerent in type with the element pointed.
For example, here is how should you define correctly a ptr:
int value = 5;
//defining a Ptr to value
int *ptr_value = &value;
//by now, ptr_value stores value's address
Anyway, you should study somewhere how this all works and how can ptrs be implemented, if you have other problems try a more specific question :)
When you are using char *h, you are declaring a pointer to a char variable. This pointer keeps the address of the variable it points to.
In simple words, as you simply declare a char variable as char num='h', then the variable num will hold the value h and so if you print it using printf("%c",num), you will get the output as h.
But, if you declare a variable as a pointer, as char *num, then it cannot actually hold any character value. I can hold only the address of some character variable.
For example look at the code below
void main()
{
char a='h';
char *b;
b=&a;
printf("%c",a);
printf("%c",b);
printf("%u",b);
}
here , we have one char variable a and one char pointer b. Now the variable a may be located somewhere in memory that we do not know. a holds the value h and &a means address of a in memory The statement b=&a will assign the memory address of a to b. Since b is declared as a pointer, It can hold the address.
The statenment printf("%c",b) will print out garbage values.
The statement printf("%u",b) will print the address of variable a in memory.
so there's difference between char num and char *num. You must first read about pointers. They are different from normal variables and must be used very carefully.
Related
My question is about dereferencing a char pointer
Here is my code -
#define MAX 10
char s[80]="Hello";
int main(){
char *stackValue;
stackValue=&s;//here I assined the address of s to stackValue
if(!stackValue){
printf("No place for Value");
exit(1);
}
else{
printf("\n%s",*stackValue);//This doesn't work with * before it
printf("\n%s",stackValue);//This works properly
}
return 0;
}
In the above code I have assigned the address of S[] to stackValue and when I am printing *stackValue it doesn't work ,
But If I print only 'stackValue' That works.
When I do same thing with Integer
int main(){
int i=10, *a;
a=&i;
printf("%d",*a);//this gives the value
printf("%d",a)//this gives the address
return 0;
}
Is printing char pointer and integer pointer is different. When I use * in int value it gives the value but gives an error when I use it as a char pointer.
Help me out?
With the first code snippet:
stackValue=&s; is incorrect given s is already an array to char. If you write like that then stackValue becomes pointer to pointer to char (not pointer to char).
Fix that by changing to stackValue=s;
Also, again %s expect a pointer to char (NOT pointer to pointer to char) - that explains why this doesn't work
printf("\n%s",*stackValue); // this doesn't work
You need printf("\n%s",stackValue); instead.
With the second code snippet.
a=&i; is ok because i is a single int, NOT an array.
What you are trying to do is this:
int main(void)
{
char a_data = "Hello, this is example";
char *pa_stack[] = {a_data};
printf("We have: %s\n", *pa_stack);
}
The "%s" format specifier for printf always expects a char* argument.
so this is working and correct statement
printf("\n%s",stackValue);
and in first statement you are passing value so it will give you undefined behaviour.
main()
{
const char **a = {"string1","string2"};
printf("%c", *a); /* prints s */
printf("%s", a); /* prints string1 */
printf("%s", a+1);/* prints ng1 */
}
GCC v4.8.3 prints "%s" for the last printf, where as http://codepad.org/ prints "ng1".
I thought that the code will create an array of pointers to two strings and the base address assigned to a, which allows normal pointer arithmetic. but it seems that there is something wrong with the assumption.The first printf suggests that my assumption is wrong. can anyone explain why this behavior is observed? ( note that VS 2012 has thrown an error saying too many initalizers where as GCC has thrown a warning for incompatible pointer assignment). I am aware of the warning due to incompatible pointer assignment.
const char **a is not an array of pointer to two strings. It declares a to be a pointer to pointer to const char.
const char **a = {"string1","string2"}; //No memory is allocated to store string literals
will invoke undefined behavior and you may get either expected or unexpected results.
To declare a as an array of two pointers you need to change the declaration as
const char *a[] = {"string1","string2"};
The memory range in your program's stack looks like this: (notice that it is not allocated before assignment which is wrong)
char** a = {s, t, r, i, n ,g, 1, \0, s, t, r, i, n, g, 2, \0}
Therefore when you print the command:
printf("%c", *a);
You are dereferencing the first character of the string which is 's'.
On the other hand, when you print the command:
printf("%s", a);
you are printing a string that starts at pointer a and finishes at '\0'. That's why you see the output 'string1'.
Lastly, when you type "a+1" you increase the pointer in one step (example here: How to increment a pointer address and pointer's value?). in this case because char** is a pointer, and every pointer is 4 byte, the "+1" jumps 4 chars forward.
Therefore when you print the command:
printf("%s", a+1);
The printf starts at the pointer 'a' + 4 bytes and ends at '\0'. That's why the output is 'ng1'.
Hope it was clear enough.
This is due to the following peculiar initialization performed by GCC.
please see int q = {1,2}; peculiar initialization list. the statement const char **a = {"string1","string2"}; results in a being treated as if const char **a = "string1". this solves the mystery as *a would print 's', a would print string1.
I'm trying to understand the mistake in the following code. The code is supposed to switch between two arrays.
What I saw is that it switches only the first 4 bytes. Is the following correct?
Passing &num1 or num1 is the same (both pass the address of the first element in the array).
The (char**) casting is wrong. That's because when you pass and array you pass the address it's laid in. So you actually pass here a void*.
How can I actually switch between these two arrays only by pointers? Is thatpossible?
I know it is possible if from the beginning I had defined char **num1 and char **num2. But I want it to stay with the array notation!
#include <stdio.h>
void fastSwap (char **i, char **d)
{
char *t = *d;
*d = *i;
*i = t;
}
int main ()
{
char num1[] = "hello";
char num2[] = "class";
fastSwap ((char**)&num1,(char**)&num2);
printf ("%s\n",num1);
printf ("%s\n",num2);
return 0;
}
Passing &num1 or num1 is the same (both pass the address of the first element in the array). Am I correct?
No. The first one is a pointer to the array itself (of type char (*)[6]), whereas in the second case, you have a pointer to the first element (of type char *; an array decays into a pointer to its first element when passed to a function).
The (char*) casting is wrong
Indeed, you are casting a char (*)[6] to a char **.
So you actualy pass here a void (Am i correct?).
No. Non sequitur. I don't see how the void type is relevant here. You have pointers, arrays, and eventually pointers to arrays.
Arrays are not pointers. Your code is trying to swap arrays, which does not make sense, since assignment to arrays is not permitted. What you probably want is
I. either get pointers to the first character of each string, and then swap the pointers themselves, like this:
void swap_pointers(const char **a, const char **b)
{
const char *tmp = *b;
*b = *a;
*a = tmp;
}
const char *p1 = "hello";
const char *p2 = "world";
swap_pointers(&p1, &p2);
II. Or use actual arrays, and you swap their contents:
void swap_contents(char *a, char *b, size_t n)
{
for (size_t i = 0; i < n; i++) {
char tmp = a[i];
a[i] = b[i];
b[i] = tmp;
}
}
char a1[] = "hello";
char a2[] = "world";
swap_contents(a1, a2, strlen(a1));
Also, you may want to read this.
1. Passing &num1 or num1 is the same (both pass the address of the first element in the array)
Not true, &num1 gives you a pointer to a pointer that points to the entire character string "hello" (char*[6]) while num1 is just a pointer to the character block "hello" (char[6]).
2. The (char*) casting is wrong. That's because When you pass and array you pass the address it's laid in. So you actualy pass here a void (Am i correct?).
It is still not a void, it's just a pointer to a character pointer. If it were void, then it would be perfectly valid to do something like void** myVoid = &num1. This will cause a syntax error unless you explicitly typecast your char** to a void** before you assign it.
The problem is your explicit type casting &num1 as a char** which is not correct, it is a char*[6]. But of course, you can't declare a variable as a char*[6] so it can't be used in this way. To fix it you need to declare your num1 and num2 as:
char* num1 = "hello";
char* num2 = "class";
instead and keep everything else the same. In fact, with this change there is no need to typecast your &num1 as a char** because it already is that.
i want to get the first character of a string (char[]) in C.
unsigned int N;
unsigned int F;
unsigned int M;
char C;
int main (int argc, char *argv[]){
if (argc!=5){
printf("Invalid number of arguments! (5 expected)\n");
exit(-1);
}
N = atoi(argv [1]);
F = atoi(argv [2]);
M = atoi(argv [3]);
C = (char) argv[4]; //this way gives a wrong char value to variable C
The code above gives me the warning: cast to pointer from integer of different size.
EDIT: as pointed in comments, argv is char *[], not char[].
There are two main ways to do this. The first is to simply dereference the pointer.
C = *argv[4];
You can also do it via array subscript, which implicitly adds an offset to the pointer before dereferencing it.
Be sure to check whether it's null first, and so on. In general, you need to be careful when dealing with pointers.
argv[4] is a char array. You need to dereference that array to obtain a single element from it
C = *(argv[4]);
I'm trying to learn C now, I'm coming from Java and there is some stuff that is new to me.
I want to print a string and send an int and a string(char array) to another method. But I keep getting some errors that I don't know how to fix.
Would really appreciate if someone could take their time and explain to me what's wrong in my code. I'm quite disoriented at the moment with these pointers. When to use %s and %c when printing etc...
Code:
#include <stdio.h>
void main()
{
int k = 10;
char string;
char *sptr;
string = "hello!";
int *ptr;
sptr = &string;
ptr = &k;
printf("%s \n", &sptr);
printf("Sending pointer.\n");
sendptr(ptr, sptr);
}
And the errors.
test.c: In function ‘main’:
test.c:8:9: warning: assignment makes integer from pointer without a cast
test.c:15:2: warning: format ‘%s’ expects type ‘char *’, but argument 2 has type ‘char **’
tezt.c: In function ‘sendptr’:
tezt.c:8:8: error: incompatible types when assigning to type ‘char[6]’ from type ‘char’
Thanks for your time! :)
First functions solved.
Second function i get this..
tezt.c: In function ‘sendptr’:
tezt.c:5:2: error: invalid initializer
#include <stdio.h>
void sendptr(int *test, char *fname)
{
char fnamn[] = &fname;
int pt;
pt = *test;
printf("%p \n", test);
printf("%d \n", pt);
printf("%s \n", fnamn);
}
char string;
string = "hello!";
First problem: you're declaring string as a single char, not as an array. Also, you can only initialize the array to a string literal in a single statement.
char string[] = "hello!";
Second problem: sptr is a pointer-to-char, so it has to point to the first element of your string. Either of these will do:
char *sptr = string;
char *sptr = &string[0];
Then, when printing the string, just pass sptr directly.
printf("%s \n", sptr);
EDIT for your next question.
char fnamn[] = &fname;
You're trying to assign a char** (pointer to pointer to char) to an array. That just won't work. If you want to copy the string pointed to by fname into fnamn then you need to use a function such as strncpy.
char fnamn[MAX_STRING_SIZE];
strncpy(fnamn, fname, MAX_STRING_SIZE);
Having said that, if you just want to print the string, then print fname directly without copying it into your array first.
Here's a corrected version of the program with some annotation:
#include <stdio.h>
int main(void) // int and (void) for standard mains.
{
int k = 10;
char *string; // a C string is a char array, you need a pointer to point to it
char *sptr;
int *ptr;
string = "hello!";
sptr = string;
ptr = &k;
printf("%s \n", sptr); // no &. The %s format expects a char*.
printf("Sending pointer.\n");
// sendptr(ptr, sptr); // don't know what this function is, ignoring
return 0;
}
In C language, the & operator means you want to use the address of the variable (ie & = "the address of the variable").
int an_integer=2; // an_integer is a memory part where you want to store 2 ;)
printf("%d", &an_integer); // here you will print the address of the memory part where an_integer is stored (not 2, more something like 2510849).
The * operator in a declaration of variable means that you want to have a pointer to a memory part, when using it in the code, it means the "the value contained at the address of"
int an_integer=2;
int *ptr_integer; // you declare a pointer to an integer
ptr_integer = &an_integer; // here you set the pointer ptr_integer to the address of an_integer
printf("%d", *ptr_integer); // here you print the value contained at the memory address stored in the ptr_integer
The [] operator means you want to store an array of something. In C, an array can be seen as a pointer to a memory space.
int an_integer[2]; // you declare an array of 2 integers
int *ptr_integer; // you declare a pointer to an integer
ptr_integer = (int *)an_integer; // here you set the value of the pointer to the address of the array, you have to cast it into an (int *) to avoid compilation warnings.
For a start, I would suggest changing:
char string;
to:
char *string;
It's pretty clear that you want the string variable to be a string rather than a single character.
In addition, you probably want to change the two lines:
sptr = &string;
printf("%s \n", &sptr);
to:
sptr = string;
printf("%s \n", sptr);
but you could equally well just pass string itself to printf.
As for the sendptr(ptr, sptr);, we can't help that much without knowing more details about it.
To fix your second function (from your edit), change:
char fnamn[] = &fname;
to:
char *fnamn = fname;
or just use fname directly. You don't have to make a copy of the pointer and the former is for things like:
char fnamn[] = "I am a string literal";
I thought it might be helpful to adding something about the difference between a char array and a pointer to a string.
In function1 below, the local variable stringPtr is a pointer to memory which contains the string "hello!". The memory containing this string will be located in a read-only section of the program. The compiler decides where to place the string "hello!" and ensures that your local variable is initialised with this memory address.
You can modify the pointer stringPtr and change it to point somewhere else. But you cannot modify the memory it points at.
Also, it is perfectly valid to use the array access notation stringPtr[2] even though it is a pointer.
In function2 the compiler will set aside 9 bytes of space on the stack for the local variable stringArray and it will ensure that this array is initialised with the string "Goodbye!". As this memory is on the stack you can modify the contents of the array.
#include <stdio.h>
void function1(void)
{
char *stringPtr = "hello!";
printf("The first char is %c\n", stringPtr[0]);
printf("The next char is %c\n", *(stringPtr+1));
// This would cause a segmentation fault, stringPtr points to read-only memory
// stringPtr[0] = 'H';
}
void function2(void)
{
char stringArray[] = "Goodbye!";
printf("The first char is %c\n", stringArray[0]);
}
int main(void)
{
function1();
function2();
return 0;
}
First of all, the return type for main should be int, not void. void main() is only well-defined if your compiler documentation explicitly lists it as a legal signature. Otherwise you invoke undefined behavior. Use int main(void) instead.
Secondly, it's time for a quick crash course on strings, arrays, and pointers.
Unlike Java, C doesn't have a dedicated string datatype; rather, strings are represented as sequences of char values terminated by a 0. They are stored as arrays of char. The string literal "hello" is stored as a 6-element array of char (const char in C++). This array has static extent, meaning it is allocated at program startup and held until the program terminates. Attempting to modify the contents of a string literal invokes undefined behavior; it's best to act as though they're unwritable.
When an array expression appears in most contexts, the type of the expression is converted from "N-element array of T" to "pointer to T", and the value of the expression is the address of the first element of the array. That's one of the reasons the string = "hello"; statement doesn't work; in that context, the type of the expression "hello" is converted from "6-element array of char" to "pointer to char", which is incompatible with the target type (which, being char, isn't the correct type anyway). The only exceptions to this rule are when the array expression is an operand of either the sizeof or unary & operators, or if it is a string literal being used to initialize another array in a declaration.
For example, the declaration
char foo[] = "hello";
allocates foo as a 6-element array of char and copies the contents of the string literal to it, whereas
char *bar = "hello";
allocates bar as a pointer to char and copies the address of the string literal to it.
If you want to copy the contents of one array to another, you need to use a library function like strcpy or memcpy. For strings, you'd use strcpy like so:
char string[MAX_LENGTH];
strcpy(string, "hello");
You'll need to make sure that the target is large enough to store the contents of the source string, along with the terminating 0. Otherwise you'll get a buffer overflow. Arrays in C don't know how big they are, and running past the end of an array will not raise an exception like it does in Java.
If you want to guard against the possibility of a buffer overflow, you'd use strncpy, which takes a count as an additional parameter, so that no more than N characters are copied:
strncpy(string, "hello", MAX_LEN - 1);
The problem is that strncpy won't append the 0 terminator to the target if the source is longer than the destination; you'll have to do that yourself.
If you want to print the contents of a string, you'd use the %s conversion specifier and pass an expression that evaluates to the address of the first element of the string, like so:
char string[10] = "hello";
char *p = string;
printf("%s\n", "hello"); // "hello" is an array expression that decays to a pointer
printf("%s\n", string); // string is an array expression that decays to a pointer
printf("%s\n", p); // p is a pointer to the beginning of the string
Again, both "hello" and string have their types converted from "N-element array of char" to "pointer to char"; all printf sees is a pointer value.
Here's a handy table showing the types of various expressions involving arrays:
Declaration: T a[M];
Expression Type Decays to
---------- ---- ---------
a T [M] T *
&a T (*)[M]
*a T
a[i] T
&a[i] T *
Declaration: T a[M][N];
Expression Type Decays to
---------- ---- ---------
a T [M][N] T (*)[N]
&a T (*)[M][N]
*a T [N] T *
a[i] T [N] T *
&a[i] T (*)[N]
*a[i] T
a[i][j] T
&a[i][j] T *
Remember that the unary & operator will yield the address of its operand (provided the operand is an lvalue). That's why your char fnamn[] = &fname; declaration threw up the "invalid initializer" error; you're trying to initialize the contents of an array of char with a pointer value.
The unary * operator will yield the value of whatever its operand points to. If the operand isn't pointing anywhere meaningful (it's either NULL or doesn't correspond to a valid address), the behavior is undefined. If you're lucky, you'll get a segfault outright. If you're not lucky, you'll get weird runtime behavior.
Note that the expressions a and &a yield the same value (the address of the first element in the array), but their types are different. The first yields a simple pointer to T, where the second yields a pointer to an array of T. This matters when you're doing pointer arithmetic. For example, assume the following code:
int a[5] = {0,1,2,3,4};
int *p = a;
int (*pa)[5] = &a;
printf("p = %p, pa = %p\n", (void *) p, (void *) pa);
p++;
pa++;
printf("p = %p, pa = %p\n", (void *) p, (void *) pa);
For the first printf, the two pointer values are identical. Then we advance both pointers. p will be advanced by sizeof int bytes (i.e., it will point to the second element of the array). pa, OTOH, will be advanced by sizeof int [5] bytes, so that it will point to the first byte past the end of the array.
#include <stdio.h>
void main()
{
int k = 10;
char string;
char *sptr;
sptr = "hello!";
int *ptr;
ptr = &k;
printf("%s \n", sptr);
printf("Sending pointer.\n");
sendptr(ptr, sptr);
}