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I'm not sure what the following does and i'm hoping someone can clarify the purpose of having the asterisk in front of the functions name:
char *Foo(char *ptr) {
return NULL;
}
I understand that you can pass by value the memory location of something in the function argument call and *ptr would be the pointer to it. I understand you can create a pointer function that can be used to point to other functions like a regular pointer points to variable memory location but in this case this is not a function pointer that we can point to other functions, or is it? This seems like a real function.
Foo is a function.
It has input: ptr of type char*
It has output of type char*
char* means "pointer to char"
it returns NULL.
That is the most plain explanation I can think of.
its misleading you, the * by the name isn't related to the name
it means the same as char* Foo(char* ptr)
which means a function which takes a char* and returns a char*
Related
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I come from a Java background and I started reading K&R, but the progress is extremely slow, because I know most of it, but still have to read everything again. So, I was thinking that I could maybe ask here some things about the C programming language to learn things a lot faster.
What I want to know is
What happens when I pass a struct variable with a &-prefix as an argument to a function? The code sample that I am trying to understand is:
struct somestruct st;
somefunction(&st);
1.1. What kind of signature does somefunction need to have and what exactly is passed?
1.2. A pointer to the struct variable would be *st instead, right?
What does it mean when a function has as a parameter sometype ** variable_name? The code that I want to understand is:
int main(int argc, char **argv)
The whole code that I want to understand is here: https://stackoverflow.com/a/35355069/3668527
Please no explanations of the code. I know what it does. I just need to know what those strange new C operators & and ** mean.
Edit: Oh, and please tell me how these operators are called!
& get pointer to that variable.
The function signature should be: void somefunction(struct somestruct *st), i.e. it will accept pointer to that structure.
strct * means pointer to strct, strct ** means pointer to pointer to strct etc.
What happens when I pass a struct variable with a &-prefix as an argument to a
function? The code sample that I am trying to understand is:
The & operator returns the adress of an object.
What kind of signature does somefunction need to have and what exactly is passed?
void somefunction(struct somestruct *pointer);
You can put in "const" in a few places to tell the compiler that you don't want to allow the method to do any changes.
1.2 A pointer to the struct variable would be *st instead, right?
Depends in which context "*st" is used, if you just want to create a pointer do it that way:
somestruct *pointer = NULL; // or init it somehow
Edit: Oh, and please tell me how these operators are called! Thanks!
They are called adress operators.
What does it mean when a function has as a parameter sometype ** variable_name
That means that the parameter is a pointer pointing to another pointer.
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I have declared a char pointer in the following manner:
School *student[10];
for(i=0;i<10;i++){
*student[i] = malloc(sizeof(Student)); <--- Error points here
}
The error I get is:
incompatible types when assigning to type 'struct Student' from type 'void*'
Does anyone know why I am getting this error?
But how come if I were to allocate memory in the same line it would be with the star. For example: Student *name = malloc(sizeof(Student)); Why does this work? Im a bit confused
*student[i] = malloc(sizeof(School)); should be student[i] = malloc(sizeof(School));
students is an array of pointer to struct of type School. So you need to allocate for each pointer in that array. When you write *student[i] - you are dereferencing pointer i instead of allocating memory for it.
And as NicolasMiari pointed out, the sizeof operator must apply to School instead of student.
But how come if I were to allocate memory in the same line it would be with the star. For example: Student *name = malloc(sizeof(Student)); Why does this work? Im a bit confused
That's different. When you write Student *name = malloc(sizeof(Student)); you are both declaring a pointer and initialize it with malloc. You can do both steps in a single line like that. Alternatively, you declare it first, then assign it with malloc in a different line - in that case you must remove the asterisk.
You may want to refer to this question pointer initialization and pointer assignment.
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it seems that this is a duplicate question, but I searched stackoverflow's question about that point and non is like to my problem(I think)
I've two variables of a struct each has its own pointer to char, when I tried to copy from one variable's string to another variable's string, nothing happened, although no errors appear, just warning
implicit declaration of function strcpy
incompatible implicit declaration of built-in function 'strcpy'
I read from some questions on stackoverflow that you'd better to use strdup() function instead of strcpy() but when I did, I had an error
too many arguments to function 'strdup'
I read that there's a problem with strcpy() called "segmentation fault" and I knew it's about memory allocation, I don't totally understand what it's exactly and don't know if it's the problem with my code?
and this is my code
struct p
{
char *name;
struct p *next;
};
struct p *ptrary[10];
int main(void)
{
struct p p,q;
p.name="xyz";
p.next=NULL;
ptrary[0]=&p;
strdup(q.name,p.name);
ptrary[1]=&q;
printf("%s\n",ptrary[1]->name);
return 0;
}
so what is the problem and how I can solve it?
strdup() takes only one argument; it malloc's and returns a new block of heap memory containing the duplicated string. (See https://stackoverflow.com/questions/252782/strdup-what-does-it-do-in-c)
Which probably also points to the problem you were having before -- were you remembering to malloc the space for q to copy p's contents into?
change
strdup(q.name,p.name);
to
q.name = strdup(p.name);
see man page of strdup for further details. The strdup() function returns a pointer to a new string.
full code:
#include <stdio.h>
#include <string.h>
struct p
{
char *name;
struct p *next;
};
struct p *ptrary[10];
int main(void)
{
struct p p,q;
p.name="xyz";
p.next=NULL;
ptrary[0]=&p;
q.name = strdup(p.name);
ptrary[1]=&q;
printf("%s\n",ptrary[1]->name);
return 0;
}
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# include <stdio.h>
# include <stdlib.h>
int main(int argc, char *argv[])
{
int a[5][10][2];
int *p;
p = (int(*)[10][2])p;//Gives error!
return EXIT_SUCCESS;
}
I want to type cast p to type so that it can act as a pointer to the given 3-d array?Is there a way to do so.I am applyindg the idea that the type of a variable is everything except variable name.
Why are you trying to "typecast" anything? Why would you expect a value "typecasted" to (int(*)[10][2]) to be compatible with an int * pointer? And why does your original code assigns p to p, completely ignoring a?
This is what you can do
int a[5][10][2];
int (*p)[10][2] = a;
Now p is a pointer that can be used to access a, i.e. p[i][j][k] is equivalent to a[i][j][k]. No typecasting necessary.
If you write p = (int(*)[10][2])a; it won't give you any errors, may be a warning. You are thinking that p will be converted to pointer to a 3-D array, which is wrong. Try this statement after assigning a to p.
printf("addresses %u %u",p,p+1);
According to you, output should be something similar to this(lets say) "addresses 9990000 99940000", because you are thinking p is pointing to 3-D array. However you will get similar to "addresses 9990000 9990004", proving that p is a pointer to an integer.
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I have a little problem with casting int to char* (string)... is it even possible in C?
I'll try to explain why i need this.
I can cast int to char but I need cast int to char*.
I had a int varriable (int number_of_revisions)
and I need convert this number of revisions to char * becouse I need create a name of file and the number of revision is part of the name.... so there is part of code for better imagination of this problem.
int number_of_revision = 970; // 970 just for example
char * version;
char * new_name;
char ch_number_of_rev[4];
version = "0.";
itoa(number_of_revision,ch_number_of_rev,10);
//strcat(version, ch_num_o_rev ); // doesn't work becouse ch_number_of_rev is char and strcat requires char*
please I need quick help... Have anybody any idea how to do it? ...
but I need cast int to char*
Casting only changes the type - it does not change the value within the variable. If you need to convert an int to array of chars (i.e. a string) then use sprintf or snprintf:
char* buffer = ... allocate a buffer ...
int value = 970;
sprintf(buffer, "%d", value);
Converting int to string in c
Also, you have not allocated any memory for version - use malloc and allocate some memory.
strcat here won't work because you haven't allocated any space to store the result in. Your version is probably in read-only memory anyway, so you'd get a segfault, otherwise you'll get memory corruption. So make sure to allocate enough space for it, e.g. by using
char version[10] = "0.";
You may want to read up on pointers first, though.