Consider the following:
#include <stdio.h>
int main()
{
int first=9;
int second=0;
double ratio;
if(second==0)
ratio="n/a";
else
ratio=(double)singularCount/pluralCount;
printf("ratio is: %f", ratio);
return 0;
}
This obviously wont run as it should.
How can I modify my code/print statement to print a float if there are no problems, and n/a if there is division by 0?
I will be printing many of these ratios in a list, so I'd like to see "n/a" when there is division by 0.
Desired Output:
ratio is: n/a
You are doing something wrong...
you shouldn't put a string into a double variable.
What you want to do is to use isnan on undefined doubles (which are completely legit to hold),
and have an if statement on it.
something like:
if (isinf(ration)) printf("ration is undefined\n");
else printf...
This way you can send the double and get it from methods/procedures/functions (that I urge you to use for code separation) and yet you'll be able to get different behaviors.
by the way - look at this code:
#include <stdio.h>
#include <math.h>
void main()
{
double d = 1.0/0.0;
if (isnan(d)) printf("it is n/a\n");
else if (isinf(d)) printf("it is inf\n");
else printf("%f\n",d);
}
another thing - %f in the case of inf or nan will just print the strings "inf" and "nan", if it's good enough for you - you don't need to change a thing, if it is for your eyes.
if (second == 0) {
printf("ratio is: n/a");
} else {
printf("ratio is: %f", (double)singularCount/pluralCount);
}
Related
I'd like a step by step explanation on how to parse the arguments of a variadic function
so that when calling va_arg(ap, TYPE); I pass the correct data TYPE of the argument being passed.
Currently I'm trying to code printf.
I am only looking for an explanation preferably with simple examples but not the solution to printf since I want to solve it myself.
Here are three examples which look like what I am looking for:
https://stackoverflow.com/a/1689228/3206885
https://stackoverflow.com/a/5551632/3206885
https://stackoverflow.com/a/1722238/3206885
I know the basics of what typedef, struct, enum and union do but can't figure out some practical application cases like the examples in the links.
What do they really mean? I can't wrap my brain around how they work.
How can I pass the data type from a union to va_arg like in the links examples? How does it match?
with a modifier like %d, %i ... or the data type of a parameter?
Here's what I've got so far:
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include "my.h"
typedef struct s_flist
{
char c;
(*f)();
} t_flist;
int my_printf(char *format, ...)
{
va_list ap;
int i;
int j;
int result;
int arg_count;
char *cur_arg = format;
char *types;
t_flist flist[] =
{
{ 's', &my_putstr },
{ 'i', &my_put_nbr },
{ 'd', &my_put_nbr }
};
i = 0;
result = 0;
types = (char*)malloc( sizeof(*format) * (my_strlen(format) / 2 + 1) );
fparser(types, format);
arg_count = my_strlen(types);
while (format[i])
{
if (format[i] == '%' && format[i + 1])
{
i++;
if (format[i] == '%')
result += my_putchar(format[i]);
else
{
j = 0;
va_start(ap, format);
while (flist[j].c)
{
if (format[i] == flist[j].c)
result += flist[i].f(va_arg(ap, flist[i].DATA_TYPE??));
j++;
}
}
}
result += my_putchar(format[i]);
i++;
}
va_end(ap);
return (result);
}
char *fparser(char *types, char *str)
{
int i;
int j;
i = 0;
j = 0;
while (str[i])
{
if (str[i] == '%' && str[i + 1] &&
str[i + 1] != '%' && str[i + 1] != ' ')
{
i++;
types[j] = str[i];
j++;
}
i++;
}
types[j] = '\0';
return (types);
}
You can't get actual type information from va_list. You can get what you're looking for from format. What it seems you're not expecting is: none of the arguments know what the actual types are, but format represents the caller's idea of what the types should be. (Perhaps a further hint: what would the actual printf do if a caller gave it format specifiers that didn't match the varargs passed in? Would it notice?)
Your code would have to parse the format string for "%" format specifiers, and use those specifiers to branch into reading the va_list with specific hardcoded types. For example, (pseudocode) if (fspec was "%s") { char* str = va_arg(ap, char*); print out str; }. Not giving more detail because you explicitly said you didn't want a complete solution.
You will never have a type as a piece of runtime data that you can pass to va_arg as a value. The second argument to va_arg must be a literal, hardcoded specification referring to a known type at compile time. (Note that va_arg is a macro that gets expanded at compile time, not a function that gets executed at runtime - you couldn't have a function taking a type as an argument.)
A couple of your links suggest keeping track of types via an enum, but this is only for the benefit of your own code being able to branch based on that information; it is still not something that can be passed to va_arg. You have to have separate pieces of code saying literally va_arg(ap, int) and va_arg(ap, char*) so there's no way to avoid a switch or a chain of ifs.
The solution you want to make, using the unions and structs, would start from something like this:
typedef union {
int i;
char *s;
} PRINTABLE_THING;
int print_integer(PRINTABLE_THING pt) {
// format and print pt.i
}
int print_string(PRINTABLE_THING pt) {
// format and print pt.s
}
The two specialized functions would work fine on their own by taking explicit int or char* params; the reason we make the union is to enable the functions to formally take the same type of parameter, so that they have the same signature, so that we can define a single type that means pointer to that kind of function:
typedef int (*print_printable_thing)(PRINTABLE_THING);
Now your code can have an array of function pointers of type print_printable_thing, or an array of structs that have print_printable_thing as one of the structs' fields:
typedef struct {
char format_char;
print_printable_thing printing_function;
} FORMAT_CHAR_AND_PRINTING_FUNCTION_PAIRING;
FORMAT_CHAR_AND_PRINTING_FUNCTION_PAIRING formatters[] = {
{ 'd', print_integer },
{ 's', print_string }
};
int formatter_count = sizeof(formatters) / sizeof(FORMAT_CHAR_AND_PRINTING_FUNCTION_PAIRING);
(Yes, the names are all intentionally super verbose. You'd probably want shorter ones in the real program, or even anonymous types where appropriate.)
Now you can use that array to select the correct formatter at runtime:
for (int i = 0; i < formatter_count; i++)
if (current_format_char == formatters[i].format_char)
result += formatters[i].printing_function(current_printable_thing);
But the process of getting the correct thing into current_printable_thing is still going to involve branching to get to a va_arg(ap, ...) with the correct hardcoded type. Once you've written it, you may find yourself deciding that you didn't actually need the union nor the array of structs.
In Java,I need to ask user input for two integers. The program needs to divide by these two integers and produce a decimal to the sixth place.
I know that i will need to name two integers: numerator and denominator. Also, I need to name a double variable: result.
Thanks for the help!!
You haven't given any clues as to the language or platform you are using, but here is a basic example written in C running on a console.
#include <stdio.h>
int main()
{
int num, den;
double quo;
printf("Enter numerator: ");
scanf("%d", &num);
printf("Enter denominator: ");
scanf("%d", &den);
if (den == 0)
printf("Divide by zero\n");
else {
quo = (double)num / (double) den;
printf("Quotient = %.6f\n", quo);
}
return 0;
}
I have an iVar named,
int DATA_IN_TRANSIT;
and I have defined several macros, e.g.
#define PLACES 0;
When I do something like the following,
if(DATA_IN_TRANSIT == PLACES)
{
NSLog(#"Make LLVM Dance!");
}
I get a compiler error (expression expected) in the line if(DATA_IN_TRANSIT == PLACES)
I don't know why it's giving me an error? Am I doing something naive?
#define PLACES 0
but without ';'
otherwise you'll get
if(DATA_IN_TRANSIT == 0;)
{
NSLog(#"Make LLVM Dance!");
}
I want to use the GSL for integration
http://www.gnu.org/software/gsl/manual/html_node/Numerical-Integration.html
However, I find no convenient way how the integrated function
(the function f in the example http://www.gnu.org/software/gsl/manual/html_node/Numerical-integration-examples.html)
can report an error to the integrator. I want to integrate a function which itself results from an integration that could fail. This is my sample program
#include <stdio.h>
#include <math.h>
#include <gsl/gsl_integration.h>
#include <gsl/gsl_errno.h>
double f (double x, void * params) {
GSL_ERROR("test error",GSL_FAILURE);
return 0.0;
}
int main (void)
{
gsl_integration_workspace * w = gsl_integration_workspace_alloc (1000);
double result, error;
gsl_function F;
F.function = &f;
gsl_set_error_handler_off();
int status = gsl_integration_qags (&F, 0, 1, 0, 1e-7, 1000,
w, &result, &error);
printf ("status = %d\n", status);
status = GSL_FAILURE;
printf ("status = %d\n", status);
gsl_integration_workspace_free (w);
return 0;
}
resulting in the output
status = 0
status = -1
I think the integrator should rather stop and return my error code. How can I achieve this?
Thank you very much for your help!!!
2011-04-27: I also tried this variant, after Brian Gough told me,
#include <stdio.h>
#include <math.h>
#include <gsl/gsl_integration.h>
#include <gsl/gsl_errno.h>
double f (double x, void * params) {
GSL_ERROR("test error",GSL_FAILURE);
return GSL_NAN;
}
int main (void)
{
gsl_integration_workspace * w = gsl_integration_workspace_alloc (1000);
double result, error;
gsl_function F;
F.function = &f;
gsl_set_error_handler_off();
int status = gsl_integration_qags (&F, 0, 1, 0, 1e-7, 1000,
w, &result, &error);
printf ("status = %d\n", status);
status = GSL_FAILURE;
printf ("status = %d\n", status);
gsl_integration_workspace_free (w);
return 0;
}
it did not help either. I will now fill out a bug report.
Thanks to Xuebin Wu from the GSL Mailing list the problem is solved:
Hi,
GSL_ERROR itself is a macro, it looks like
gsl_error (reason, __FILE__, __LINE__, gsl_errno);
return gsl_errno;
The function already returns before you return NAN, because GSL_ERROR
has been called. Turning the handler off just let the first line do
nothing. The default error handler abort the program after printing
error message.
I do not think it is a bug. Maybe you can write your own error handler
to solve your problem. For example, you can use "goto" to jump out of
gsl_integration_qags, or set some global variable to indicate the
integration result is incorrect.
PS: I believe this macro is what you need,
Macro: GSL_ERROR_VAL (reason, gsl_errno, value)
This macro is the same as GSL_ERROR but returns a user-defined value
of value instead of an error code. It can be used for mathematical
functions that return a floating point value.
The following example shows how to return a NaN at a mathematical
singularity using the GSL_ERROR_VAL macro,
if (x == 0)
{
GSL_ERROR_VAL("argument lies on singularity",
GSL_ERANGE, GSL_NAN);
}
So I adjusted the code according to
#include <stdio.h>
#include <math.h>
#include <gsl/gsl_integration.h>
#include <gsl/gsl_errno.h>
double f (double x, void * params) {
// return GSL_NAN;
GSL_ERROR_VAL ("argument lies on singularity", GSL_ERANGE, GSL_NAN);
}
int main (void)
{
gsl_integration_workspace * w = gsl_integration_workspace_alloc (1000);
double result, error;
gsl_function F;
F.function = &f;
gsl_set_error_handler_off();
int status = gsl_integration_qags (&F, 0, 1, 0, 1e-7, 1000,
w, &result, &error);
printf ("status = %d\n", status);
status = GSL_FAILURE;
printf ("status = %d\n", status);
gsl_integration_workspace_free (w);
return 0;
}
and everything works as expected...
A bit hackish, but I'd probably have your function store some flag. When it encounters an error it sets the flag and returns zero for all subsequent evaluations. Then, after you've integrated it you can check this flag to see if the result is valid.
What about to write a wrapper for the function which returns pointer to a structure, containing function results and error status ? Or if you use c++, this encapsulation can be made with use of objects ....
Good afternoon,
I'm trying to accomplish a task that i know should be doable. however my attempts seem to fail every time. My endeavor is to learn to code in Objective -c and have been making good progress. what i would like to do is add a loop to my current application that asks at the end if i would like to run again or some thing to that regard, and reply with a yes or no. if no the program ends and if yes it jumps back to the top of the project to start all over. kinda like what i have below? forgive me please if its not quite perfect, im still getting used to programing and am finding it incredibly fun.
#include <stdio.h>
int main(void)
{
char loop = yes;
while (loop = yes)
{
.
.
.
}
printf ("would you like to continue (yes/no)/n");
scanf ("%s", loop);
}
The printf and scanf need to be moved up inside the curly braces of the while loop. Also, you want \n instead of /n in the printf. Finally, you're going to get a string back with that scanf() call, so you'll want to declare loop as a char array, and then in the while loop, check the first element of that array for a 'y' or 'n' or something like that. You might also want to look at getchar() instead of scanf() for that sort of thing.
Not compiled here, but should work:
#include <stdio.h>
int main(void)
{
char buffer[256];
do {
.
.
.
printf ("would you like to continue (yes/no)/n");
scanf ("%s", buffer);
} while (strcmp(buffer,"yes") != 0);
}
One wouldn't do anything like that in a real world application, but for demonstration purpose it should be ok.
I made your variable an array, because strings are arrays of characters in C. Length is set to 256 bytes (255 characters + 0-byte as delimiter). I changed the loop to do-while to make it run at least once. For string comparison you need to call a function. strcmp returns 0 for identical strings. Finally, the question belongs in the loop.
It is plain C though, using nothing of Objective-C.
int main() {
char A = 'n';
char B = 'y';
char Answer;
printf("Does the subject have a glazed over look? (y/n): \n");
scanf("%c",&Answer);
if (Answer=='N'||Answer=='y'|| Answer=='N'||Answer=='Y')
printf("Good\n");
else
printf("Please enter 'y' or 'n' \n ");
return 0;
}
#include <stdio.h>
int main(void)
{
avi;
char loop[10];
while (loop = yes)
{
.
.
.
}
printf ("would you like to continue (yes/no)/n");
scanf ("%s", loop);
if(strcpm(loop,"YES")==0) goto avi:
}