Ok so, I'm trying to write a program which numerically evaluates integrals using Simpson's 3/8 rule. I'm having issues passing the values from Integral *newintegral to the simpson() function. I'm not massively confident in my understanding of structures and pointers, and I've been reviewing the lecture notes and checking online for information all day and I still can't understand why it's not working.
At the moment when I try to build my program it comes up with a number of errors, particularly: on line 46 "expected expression before Integral" and on most of 55-63 "invalid type of argument of '->' (have 'Integral') I don't understand why the first one is occurring because all my lecturers examples of this type of thing, when passing a structure to a function just have the syntax func(Struct_define_name individual_struct_name). I thought this is what I was doing with mind (Integral being the name of the structure type and i being the specific structure) but obviously not.
I think these two problems are connected so I included all of my code for context, however the lines which actually have errors are 46 and 55-63 as mentioned above. I've probably defined the structure wrong in the first place or something though.
(Incidentally the maths in the simpson() function doesn't actually work properly now anyway, but that's not something I'm concerned about)
Also I tried looking at other similar questions but I didn't understand what the other code was doing so I couldn't extrapolate how to fix my code from that. I know this isn't very relevant to other people but I really don't understand programming well enough to try and phrase my question in a general sense...
'#include <stdio.h>
#include <stdlib.h>
#include <math.h>
typedef struct integral {
double result, limits[2];
int degree;
double coefficients[];
} Integral;
// Prototype of function that calculates integral using Simpson's 3/8 rule
double simpson(Integral i);
// function (?) which initialises structure
Integral *newintegral() {
Integral *i = malloc(sizeof *i);
double lim1_in, lim2_in;
int degree_input, n;
printf("Please enter the degree of the polynomial.\n");
scanf("%d", °ree_input);
i->degree = degree_input;
printf("Please enter the %d coefficients of the polynomial, starting\n"
"from the highest power term in the polynomial.\n", (i->degree+1));
for (n=i->degree+1; n>0; n=n-1) {
scanf("%lg", &i->coefficients[n-1]);
}
printf("Please enter the upper limit of the integral.\n");
scanf("%lg", &lim1_in);
i->limits[0] = lim1_in;
printf("Please enter the lower limit of the integral.\n");
scanf("%lg", &lim2_in);
i->limits[1] = lim2_in;
return i;
}
int main() {
Integral *i = newintegral();
simpson(Integral i);
return 0;
}
double simpson(Integral i) {
int n;
double term1, term2, term3, term4;
for (n=(i->degree); n>0; n=n-1) {
term1=(pow(i->limits[1],n)*(i->coefficients[n]))+term1;
term2=(pow(((((2*(i->limits[1]))+(i->limits[0])))/3),n)*(i->coefficients[n]))+term2;
term3=(pow(((((2*(i->limits[0]))+(i->limits[1])))/3),n)*(i->coefficients[n]))+term3;
term4=(pow(i->limits[0],n)*(i->coefficients[n]))+term4;
}
i->result = (((i->limits[0])-(i->limits[1]))/8)*(term1+(3*term2)+(3*term3)+term4);
printf("The integral is %lg\n", i->result);
return 0;
}'
You're currently passing a pointer to a function that takes a single Integral argument.
Your prototype, double simpson(Integral i); tells the compiler "declare a function called simpson that returns a double and takes a single Integral referenced by the identifier i inside the function.
However, in main() you say:
int main() {
//declare a pointer to an Integral and assign it to the return of 'i'
Integral *i = newintegral();
//call the function simpson with i.
//However, you are redeclaring the type of the function argument, so the compiler will complain.
simpson(Integral i);
return 0;
}
Your call, simpson(Integral i); will not work because you are redeclaring the type of the function argument. The compiler will state:
:46:13: error: expected expression before ‘Integral’
What you really need is for simpson() to take a pointer to Integral as its argument. You have actually already handled this inside the function, (using i->) but your function prototype is telling the compiler that you are passing the whole struct Integral as the function argument.
Solution:
Change your function prototype as follows:
double simpson(Integral *i); // function returning double taking single pointer to an Integral named i.
...and change main() to look like the following:
int main(void) { //In C main has two valid definitions:
//int main(void), or int main(int argc, char **argv)
Integral *i = newintegral();
simpson(i);
return 0;
}
So in conclusion, your understanding of pointers is correct, but not how you pass a pointer to a function.
**Sidenote:
Remember to always build your code with all warnings enabled. The compiler will give you very useful diagnostics that will help you quickly find solutions to problems like this. For GCC, as a minimum, use gcc -Wall myprogram.c
Two obvious problems:-
Line 46 : simpson(Integral i);
...should be just simpson(i);. Putting a type there is simply an error.
And this, later:
double simpson(Integral i)
.. tells the compiler to pass in Integral object yet you use the indirection operator i.e i->limits as though you'd been passed a pointer. The easiest fix is to make the function expect a pointer, like this:
double simpson(Integral *i)
Related
Here's a code:
#include <stdio.h>
/* declaration */
int square (int num);
int main() {
int x, result;
x = 5;
result = square(x);
printf("%d squared is %d\n", x, result);
return 0;
}
/* definition */
int square (int num) {
int y;
y = num * num;
return(y);
}
And when I change it like this:
#include <stdio.h>
/* declaration */
int main() {
int x, result;
x = 5;
result = square(x);
printf("%d squared is %d\n", x, result);
return 0;
}
/* definition */
int square (int num);
int square (int num) {
int y;
y = num * num;
return(y);
}
I get the warning:
Implicit declaration of the function 'square'
Does this warning has any relation with the branch prediction things? Does the former code improves branch prediction and hence the performance of the code (as the function call branches/diverts the execution from the normal execution of the code!?)
No, it has nothing to do with branch prediction.
The language requires that a declaration of a function appear before the function is called. The compiler needs to see the declaration in order to generate code for the function call that is even correct, let alone performant: to know what types are expected as parameters, what type is returned, how the parameters should be laid out in the stack and registers, and so on. And by the one-pass design of the C language, this means the declaration has to appear before the function call. It's true that the necessary information is further down in the source file, but that's too late - the compiler isn't designed to "peek ahead" for it.
In modern C, the absence of the declaration makes the program invalid, and the compiler must issue a diagnostic. Previous versions of the C language allowed such a program to compile, using an "implicit declaration" which was essentially a standard set of rules to guess what types were expected and returned, and how they should be passed. Some compilers, such as gcc, proceed under these earlier rules when they find a missing declaration, and only give a warning instead of an error (which complies with the requirement for a diagnostic). But as the programmer, you really should consider it an error, since there's a good chance that the "guessed" implicit declaration will be wrong for your function, in which case the compiled program may fail in unpredictable ways at runtime.
The code given below is an exercise that our teacher gave to prepare us for exams.
We are supposed to find the errors that occur in this code and fully explain them .
#define SIZE 10
int start (void a,int k) {
const int size=10;
char array[size];
char string[SIZE];
mycheck(3,4);
array[0]=string[0]='A';
printf("%c %c\n", array[0], string[0]);
myRec(7);
}
int mycheck(int a , int b) {
if (a==0 || b==0 ) {
return 0;
}
else {
return (a*b);
}
}
int myRec(int x) {
if(x==0)
return 0;
else
printf("%d,",x);
myRec(x--);
}
I have found these errors so far:
1.int start (void a,int k)
explanation: We can't have a variable of type void, because void is an incomplete type
2.const int size=10;
explanation:we can't use variable to define size of array
(problem is when I run it in dev-c++ it doesn't show an error so I'm not sure about this)
3.mycheck(3,4);
explanation: prototype of function mycheck() is not declared, so the function mycheck is not visible to the compiler while going through start() function
4.A friend told me that there is an error in function myRec because of this statement myRec(x--);
(I don't really get why is this an error and how you can I explain it?)
5.Main() function doesn't exist.
I'm not sure about this but if i run the code (in dev-c++) without main function I get a compilation error
I'm not sure if the errors that I pointed out are 100% right or if I missed an error or if I explained them correctly.
Please correct me if any of the above is wrong!
a friend told me that there is an error in function myRec cuz of this
statement myRec(x--);
It will lead to stackoverflow. Due to post-decrement, the actual argument passed to function myRec(), never decreases and therefore the condition:
if(x==0)
return 0;
will never become true. Regarding your rest of the errors, it depends on the compiler version being used:
For example C99, you are allowed to have variable size arrays like this:
const int size=10;
char array[size];
char string[SIZE];
but pre C99, you would have to use malloc or calloc. For your functions used without prototype, most compilers would generate a warning and not error and also due to no #include<stdio.h> statement, your printf would also lead to a warning.i Again, lot of these things are compiler dependent.
1.int start (void a,int k)
explanation: We can't have a variable of type void ,because void is an
incomplete type
Correct.
2.const int size=10;
explanation:we can't use variable to define size of array (problem is
when i run it in dev-c++ it doesnt show an error?so im not sure about
this!)
This is also correct, that char array[size];, where size is not a compile-time constant, is invalid in C89. However, in C99 and newer, this is actually valid and would create a variable-length array. It is possible that your Dev-C++ IDE is using GCC with the language set to C99 or newer, or has GNU C extensions enabled to enable this feature.
3.mycheck(3,4);
explanation: prototype of function mycheck() is not declared.So the
function mycheck is not visible to the compiler while going through
start() function
Correct. This can be fixed either by declaring the function's prototype before the start() function, or just moving the whole function to the top of the file. As noted by Toby Speight in the comments, in C89, this should not actually be a compiler error, since functions are implicitly declared when they are used before any actual declaration as int (), i.e. a function returning int with any arguments, which is compatible with the declarations of mycheck and myRec. It is however bad practice to rely on this, and implicit function declaration does not work in C99 or newer.
4.a friend told me that there is an error in function myRec cuz of this statement myRec(x--);
(I don't really get why is this an error and how you can explain it?)
This function is a recursive function. This means it calls itself within itself in order to achieve a kind of looping. However, this function as it is currently written would run forever and cause an infinite loop, and since it is a recursive function, and needs a new stack frame each time it is called, it will most likely end in a stack overflow.
The function is written with this statement:
if(x==0)
return 0;
This is intended to terminate the recursion as soon as x reaches 0. However, this never happens, because of this line of code here:
myRec(x--);
In C, postfix -- and ++ operators evaluate to their original value before the addition or subtraction:
int x = 5;
int y = x--;
/* x is now 4; y is now 5 */
However, using the prefix version of these operators will evaluate to their new value after adding / subtracting 1:
int x = 5;
int y = --x;
/* x is now 4; y is now 4 */
This means that on each recursion, the value of x never actually changes and so never reaches 0.
So this line of code should actually read:
myRec(--x);
Or even just this:
myRec(x - 1);
5.Main() function doesn't exist ...again im not sure about this but if i run the code (in dev-c++) without main function i get a compilation
error
This one could either be right or wrong. If the program is meant to run on its own, then yes, there should be a main function. It's possible that the function start here should actually be int main(void) or int main(int argc, char *argv[]). It is entirely valid however to compile a C file without a main, for example when making a library or one individual compilation unit in a bigger program where main is defined in another file.
Another problem with the program is that myRec is used before it is declared, just like your point 3 where mycheck is used before it is declared.
One more problem is that the functions start and mycheck are declared to return int, yet they both do not contain a return statement which returns an int value.
Other than that, assuming that this is the entire verbatim source of the program, the header stdio.h isn't included, yet the function printf is being used. Finally, there's the issue of inconsistent indentation. This may or may not be something you are being tested for, but it is good practice to indent function bodies, and indentation should be the same number of spaces / tab characters wherever it's used, e.g.:
int myRec(int x) {
if(x==0)
return 0;
else
printf("%d,",x);
myRec(x--);
}
1) Hello friend your Recursive function myRec() will go infinite because it
call itself with post detriment value as per C99 standard it will
first call it self then decrements but when it call itself again it have
to do the same task to calling self so it will never decrements and new
stack is created and none of any stack will clear that recursion so
stack will full and you will get segmentation fault because it will go
beyond stack size.
2) printf("%d,",x); it should be printf("%d",x); and you should include #include library.
I think your another mistake is you are calling your mycheck() and you
returning multiplication of two integer but you are not catch with any
value so that process got west.So while you are returning something you
must have to catch it otherwise no need to return it.
3) In this you Program main() function missing. Program execution starts
with main() so without it your code is nothing. if you want to execute
your code by your own function then you have to do some process but
here main() should be present.or instead of start() main() should
be present.
4) you can also allocate any char buffer like this int j; char array[j=20];
your code should be like this.
#include<stdio.h>
#define SIZE 10
int mycheck(int a , int b) {
if (a==0 || b==0 ) {
return 0;
}
else {
return (a*b);
}
}
int myRec(int x) {
if(x==0)
return 0;
else
printf("%d",x);
myRec(--x);
}
void main (int argc, char** argv) {
const int size=10;
char array[size];
char string[SIZE];
int catch = mycheck(3,4);
printf("return value:: %d\n",catch);
array[0]=string[0]='A';
printf("%c %c\n", array[0], string[0]);
myRec(7);
printf("\n");
}
Enjoy.............
I have created an array of function pointers to swap two variables.
pointer pointing to these functions namely: swap1, swap2. swap3 and swap4.
swap2 is swaping using pointer passed as arguments.
but while declaring the function pointer, only int and int are passed as arguments. after compiling this causes many warnings.
so do we have a better way of passing the argument, where we put condition in function call itself.
code is given below.
#include <stdio.h>
int swap1(int ,int );
int swap2(int* ,int* );
int swap3(int ,int );
int swap4(int, int);
int swap1(int a,int b)
{
int temp=a;
a=b;
b=temp;
printf("swapped with 3rd variable :%d, %d\n", a,b);
}
int swap2(int *a,int *b)
{
int temp = *a;
*a = *b;
*b = temp;
printf("swapped with pointer :%d, %d\n", *a,*b);
}
int swap3(int a,int b)
{
a+=b;
b=a-b;
a-=b;
printf("swapped with 2 variable :%d, %d\n", a,b);
}
int swap4(int a,int b)
{
a=a^b;
b=a^b;
a=a^b;
printf("swapped with bitwise operation :%d, %d\n", a,b);
}
int main()
{
int ch;
int a=3;
int b=4;
printf("enter the option from 0 to 3\n");
scanf("%d",&ch);
int (*swap[4])(int, int) ={swap1,swap2,swap3,swap4};// function pointer
/*can we pass something like int(*swap[4]( condition statement for 'pointer to variable' or 'variable')*/
if (ch==1)// at '1' location, swap2 is called.
{
(*swap[ch])(&a,&b);//passing the addresses
}
else
{
(*swap[ch])(a,b);
}
return 0;
}
some warnings are as follows.
at line 36 in file '9e748221\script.c'
WARNING: found pointer to int where int is expected
at line 47 in file '9e748221\script.c'
WARNING: found pointer to int where int is expected
at line 47 in file '9e748221\script.c'
Well yes. There are a number of problems with your code, but I'll focus on the ones to which the warnings you presented pertain. You declare swap as an array of four pointers to functions that accept two int arguments and return an int:
int (*swap[4])(int, int)
Your function swap2() is not such a function, so a pointer to it is not of the correct type to be a member of the array. Your compiler might do you a better favor by rejecting the code altogether instead of merely emitting warnings.
Having entered a pointer to swap2() into the array anyway, over the compiler's warnings, how do you suppose the program could call that function correctly via the pointer? The type of the pointer requires function arguments to be ints; your compiler again performs the dubious service of accepting your code with only warnings instead of rejecting it.
Since the arguments in fact provided are the correct type, it might actually work on systems and under conditions where the representations of int and int * are compatible. That is no excuse, however, for writing such code.
Because pointers and ints are unchanged by the default argument promotions, one alternative would be to omit the prototype from your array declaration:
int (*swap[4])() = {swap1,swap2,swap3,swap4};
That says that each pointer points to a function that returns int and accepts a fixed but unspecified number of arguments of unspecified types. At the point of the call, the actual arguments will be subject to the default argument promotions, but that is not a problem in this case. This option does prevent the compiler from performing type checking on the arguments, but in fact you cannot do this correctly otherwise.
Your compiler might still warn about this, or could be induced to warn about it with the right options, but the resulting code nevertheless conforms and does the right thing, in the sense that it calls the pointed-to functions with the correct arguments.
To deal with the warnings first: You declare an array of functions which take int parameters. This means that swap2 is incompatible with the type of element for the array you put it in. This will generate a diagnostic.
Furthermore, when you call one of the functions in the array, the same array declaration tells the compiler that the parameters need to be ints not pointers to int. You get two diagnostics here, one for each parameter.
To fix the above all your functions need to have compatible prototypes with the element type of the array. Should it be int or int*? This brings us to the other problem.
C function arguments are always pass by value. This means that the argument is copied from the variable onto the stack (or into the argument register depending on the calling convention and argument count - for the rest of this post, I'll assume arguments are placed on the stack for simplicity's sake). If it's a literal, the literal value is put on the stack. If the values on the stack are changed by the callee no attempt is made by the caller, after the function returns, to put the new values back in the variables. The arguments are simply thrown away.
Therefore, in C, if you want to do the equivalent of call by reference, you need to pass pointers to the variables you use as arguments as per swap2. All your functions and the array should therefore use int*. Obviously, that makes one of swap1 and swap2 redundant.
The correct array definition is
int (*swap[4])(int*, int*) = {swap1, swap2, swap3, swap4};
and the definition of each function should be modified to take int* parameters. I'd resist the temptation to use int (*swap[4])() simply because it circumvents type safety. You could easily forget the & in front of an int argument when the called function is expecting a pointer which could be disastrous - the best case scenario when you do that is a seg fault.
The others have done great work explaining what the problems are. You should definitely read them first.
I wanted to actually show you a working solution for that sort of problem.
Consider the following (working) simple program :
// main.c
#include <stdio.h>
void swap1(int* aPtr, int* bPtr) {
printf("swap1 has been called.\n");
int tmp = *aPtr;
*aPtr = *bPtr;
*bPtr = tmp;
}
void swap2(int* aPtr, int* bPtr) {
printf("swap2 has been called.\n");
*aPtr += *bPtr;
*bPtr = *aPtr - *bPtr;
*aPtr -= *bPtr;
}
int main() {
int a = 1, b = 2;
printf("a is now %d, and b is %d\n\n", a, b);
// Declare and set the function table
void (*swapTbl[2])(int*, int*) = {&swap1, &swap2};
// Ask for a choice
int choice;
printf("Which swap algorithm to use? (specify '1' or '2')\n>>> ");
scanf("%d", &choice);
printf("\n");
// Swap a and b using the right function
swapTbl[choice - 1](&a, &b);
// Print the values of a and b
printf("a is now %d, and b is %d\n\n", a, b);
return 0;
}
First of, if we try to compile and execute it:
$ gcc main.c && ./a.out
a is now 1, and b is 2
Which swap algorithm to use? (specify '1' or '2')
>>> 2
swap2 has been called.
a is now 2, and b is 1
As myself and others mentioned in answers and in the comments, your functions should all have the same prototype. That means, they must take the same arguments and return the same type. I assumed you actually wanted to make a and b change, so I opted for int*, int* arguments. See #JeremyP 's answer for an explanation of why.
I have to learn it for my study. Is their any way to cast a pointer to integer. I have to give myEulerForward1 a pointer as a paramter and i always get this error message :
eulerZahl.c: In function ‘main’:
eulerZahl.c:38:35: warning: passing argument 1 of ‘myEulerForward1’ makes pointer from integer without a cast [-Wint-conversion]
double forward = myEulerForward1(k);
^
eulerZahl.c:16:8: note: expected ‘int *’ but argument is of type ‘int’
double myEulerForward1(int *n1){
^~~~~~~~~~~~~~~
Can someone help me with it?
#include <stdio.h>
#include <stdint.h>
double kFactorial(int k){
if(k <= 1){
return 1;
}
return k * kFactorial(k - 1);
}
double myEulerForward1(int *n1){
double n = 1;
double euler, nFact = 1;
for(int i = sizeof(&n1); i >= 0; i--){
nFact*= i;
euler = euler + (1.0/nFact);
}
return euler;
}
int main(){
int k = 4;
double factorial = kFactorial(k);
printf("The factorial of %d is: %lf ", k, factorial);
double forward = myEulerForward1(k);
printf("The Eulers Number: %lf", forward);
}
First i can see an error at sizeof(&n1);
Of course n1 is a pointeur. His Value is a RAM address. But for see his Deferenced value, you must use * before n1. & is used for get the adress of something. * is used for get the inside of a pointer.
Use : sizeof(*n1);
In Second, i see that you get a pointer in the prototype of myEulerForward1
double myEulerForward1(int *n1)
It's your compile error. He said that your function need a pointer (an adress) and that you put everything except that.
So when you call this function, you must put a pointer (a RAM adress).
And for do that, in the calling of the function, you must use & of course for get the adress of n1 and not his number value.
Use : double forward = myEulerForward1(&k);
Well, yes, it is possible to convert a pointer to an int, and vice versa.
However, you are making a serious mistake in asking that question. There are circumstances where converting an int to a pointer, or vice versa, makes sense. But, in your code, you would be using it as a blunt instrument to force the compiler to accept bad code.
Your compiler is complaining because you have passed an int to a function that expects an int *.
Forcing the issue, by converting that int to a pointer, will stop the compiler complaining, but then you'll (possibly) get some form of runtime error, since the function will receive an invalid pointer.
Your choices are
remove the * from double myEulerForward1(int *n1). This will mean, the function expects an int, so your code that passes an int will be correct.
Call the function as myEulerForward1(&k) which passes an address of k (which is not k converted to a pointer) as a pointer.
Looking at the body of myEulerForward1() there are other problems as well. You need to read up and better understand what sizeof does. Whether your function accepts an int or a pointer (int *) the logic of your function is faulty.
I wrote a simple C program because I just got started learning how to program.
Here is the main.c file:
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
float mul(float r[],float o[]){
o[0] = r[0]*11;
o[1] = r[1]*22;
o[2] = r[2]*33;
}
//==============================================================
int main(void){
float r[3];
r[0]=1;r[1]=2;r[2]=3;
float o[3];
o=mul(r,o);
return 0;
}`
My goal is to fill vector o by using the function mul; I am required to keep the definition of the variables as it is; the only thing that I am supposed to modify is the function.
When I run it by typing gcc main.c I get an error message like this:
error: incompatible types when assigning to type ‘float[3]’ from type ‘float’
o=mul(r,o);
And I have no Idea how to fix it. What am I supposed to change in my little code? as far as I learned I should pass to the function the pointers referred to the vectors but it should be the same thing of what I have done since the vector name is the pointer to the vector itself.
Thanks for your help
In C it's called an "array", not a "vector".
You can't assign a value to an array.
You have mul declared as returning a float, but it doesn't return anything - change that to void.
When you call mul, since it doesn't return anything, you don't need the o=. Just mul(r,o), will be fine.
An array is not a pointer. That said, an array does decay into a pointer to its first element in most contexts, including the function call you're using it in. Likewise, the float r[], float o[] in your mul function signature is just syntactic sugar for float *r, float *o.
o=mul(r,o);
assigns o to the return value of the mul function. But mul has no return value--it has no return statement at all, nor does it need one, because it operates directly on the arrays passed to it. Instead, just declare it void:
void mul(float r[], float o[]) {
. . .
Then call it without the assignment:
mul(r, o);