int a;
printf("%d\n", a);
I wonder if %d is a cast?
In any case it won't be a cast but a reinterpretation (like getting the address, casting to a pointer of a different type and then getting the contents as a new type).
Example:
printf("%d\n", 1.5);
won't print integer 1, but the integer value of the representation of 1.5 in IEEE 754. If you want to cast, you must explicitly put (int) in front of the value.
No, it is part of the format specifier string for printf() function's first argument; the format string. It will print out a decimal representation of that int you passed as the second argument.
It is not. It is just a "hint" for printf() function to treat the 'a' argument as an 'int'
No, it's not a cast, but I suggest you take a look at the source for printf() to understand this. There's nothing special about printf() -- it's just a varargs function like any other. It's one of the first functions you learn in C, usually well before you learn varargs, so it often sticks out in people's minds as special when it's really not. A quick study of the source will probably be enlightening.
When you pass a format string to printf(), you're telling the function what to expect in its argument list (generally on the stack), but that might not agree with what you actually put there. With %d, you're telling printf() to take the next integer-sized chunk of bytes off the argument list and format those bytes as if they represent a signed decimal number. So when printf() parses the format string and encounters a %d, it will probably do something like:
int num = va_arg(args, int);
And then format and output the bytes in "num" as if they were an integer, regardless of what kind of argument you actually passed. If you put a float in the arguments where printf() is told to expect an integer, the output will be a decimal representation of the IEEE floating point bytes -- probably not what you intended, and not what a true cast would have done.
No, it is a format specifier. It has semantic meaning only to the formatted I/O functions and is not part of the C language itself. You could equally write yopur own
All it does is specify the 'human readable' representation in which to present an int value; there is no type conversion or translation.
Related
void main()
{
clrscr();
float f = 3.3;
/* In printf() I intentionaly put %d format specifier to see
what type of output I may get */
printf("value of variable a is: %d", f);
getch();
}
In effect, %d tells printf to look in a certain place for an integer argument. But you passed a float argument, which is put in a different place. The C standard does not specify what happens when you do this. In this case, it may be there was a zero in the place printf looked for an integer argument, so it printed “0”. In other circumstances, something different may happen.
Using an invalid format specifier to printf invokes undefined behavior. This is specified in section 7.21.6.1p9 of the C standard:
If a conversion specification is invalid, the behavior is
undefined.282) If any argument is not the correct type for the
corresponding conversion specification, the behavior is undefined.
What this means is that you can't reliably predict what the output of the program will be. For example, the same code on my system prints -1554224520 as the value.
As to what's most likely happening, the %d format specifier is looking for an int as a parameter. Assuming that an int is passed on the stack and that an int is 4 bytes long, the printf function looks at the next 4 bytes on the stack for the value given. Many implementations don't pass floating point values on the stack but in registers instead, so it instead sees whatever garbage values happen to be there. Even if a float is passed on the stack, a float and an int have very different representations, so printing the bytes of a float as an int will most likely not give you the same value.
Let's look at a different example for a moment. Suppose I write
#include <string.h>
char buf[10];
float f = 3.3;
memset(buf, 'x', f);
The third argument to memset is supposed to be an integer (actually a value of type size_t) telling memset how many characters of buf to set to 'x'. But I passed a float value instead. What happens? Well, the compiler knows that the third argument is supposed to be an integer, so it automatically performs the appropriate conversion, and the code ends up setting the first three (three point zero) characters of buf to 'x'.
(Significantly, the way the compiler knew that the third argument of memset was supposed to be an integer was based on the prototype function declaration for memset which is part of the header <string.h>.)
Now, when you called
printf("value of variable f is: %d", f);
you might think the same thing happens. You passed a float, but %d expects an int, so an automatic conversion will happen, right?
Wrong. Let me say that again: Wrong.
The perhaps surprising fact is, printf is different. printf is special. The compiler can't necessarily know what the right types of the arguments passed to printf are supposed to be, because it depends on the details of the %-specifiers buried in the format string. So there are no automatic conversions to just the right type. It's your job to make sure that the types of the arguments you actually pass are exactly right for the format specifiers. If they don't match, the compiler does not automatically perform corresponding conversions. If they don't match, what happens is that you get crazy, wrong results.
(What does the prototype function declaration for printf look like? It literally looks like this: extern int printf(const char *, ...);. Those three dots ... indicate a variable-length argument list or "varargs", they tell the compiler it can't know how many more arguments there are, or what their types are supposed to be. So the compiler performs a few basic conversions -- such as upconverting types char and short int to int, and float to double -- and leaves it at that.)
I said "The compiler can't necessarily know what the right types of the arguments passed to printf are supposed to be", but these days, good compilers go the extra mile and try to figure it out anyway, if they can. They still won't perform automatic conversions (they're not really allowed to, by the rules of the language), but they can at least warn you. For example, I tried your code under two different compilers. Both said something along the lines of warning: format specifies type 'int' but the argument has type 'float'. If your compiler isn't giving you warnings like these, I encourage you to find out if those warnings can be enabled, or consider switching to a better compiler.
Try
printf("... %f",f);
That's how you print float numbers.
Maybe you only want to print x digits of f, eg.:
printf("... %.3f" f);
That will print your float number with 3 digits after the dot.
Please read through this list:
%c - Character
%d or %i - Signed decimal integer
%e - Scientific notation (mantissa/exponent) using e character
%E - Scientific notation (mantissa/exponent) using E character
%f - Decimal floating point
%g - Uses the shorter of %e or %f
%G - Uses the shorter of %E or %f
%o - Signed octal
%s - String of characters
%u - Unsigned decimal integer
%x - Unsigned hexadecimal integer
%X - Unsigned hexadecimal integer (capital letters)
%p - Pointer address
%n - Nothing printed
The code is printing a 0, because you are using the format tag %d, which represents Signed decimal integer (http://devdocs.io).
Could you please try
void main() {
clrscr();
float f=3.3;
/* In printf() I intentionaly put %d format specifier to see what type of output I may get */
printf("value of variable a is: %f",f);
getch();
}
This question already has answers here:
What happens when I use the wrong format specifier?
(2 answers)
Correct format specifier to print pointer or address?
(5 answers)
Closed 4 years ago.
As you can see in the program, The first output is 6356744 and the second output is 0060FF08, why is it different? Is the %d typecasting it into an integer, if so, how?
#include<stdio.h>
int main()
{
int *a;
int b = 7;
a = &b;
printf(" The value of a = %d",a);
printf("\n The value of a= %p",a);
}
Printing a pointer with %d is formally undefined behavior, meaning anything can happen, including a program crash. Your program will for example likely break when you compile it as a 64 bit application, where int is 32 bits but a pointer is likely 64 bits. Therefore, always use %p and never anything else when printing a pointer.
There is no implicit conversion taking place - the printf family of functions doesn't have that kind of intelligence - it doesn't know the type passed. With the format specifier, you tell the function which type it is getting. And if you lie to printf and say "I'll give you an int" and then give it a pointer, you unleash bugs. This makes the printf family of functions very dangerous in general.
(The only implicit conversion that take place in printf is when you pass small integer types or float, in which case the "default argument promotions" take place and promote the parameter either to int or double. This is not the case here, however.)
In this specific case, you happened to get the decimal representation of 0x0060FF08, which is by no means guaranteed.
Pedantically, you should also cast the pointer to type (void*) since this is what %p expects.
%p prints a pointer and it's not necessarily hexadecimal, or even a number
If format specifiers do not match the datatype of the provided parameter, you yield undefined behaviour. %d expects an integral value, such that when you pass a pointer value, you get undefined behaviour (cf., for example, cppreference.com-printf):
...If any argument after default argument promotions is not the type
expected by the corresponding conversion specifier, or if there are
fewer arguments than required by format, the behavior is undefined.
The (only) correct format specifier for printing pointer values is %p, usually printing the address in hex format.
One of the undefined behaviours is that %d takes the pointer value as an 32/64 bit integral value and hence prints decimals, which - if you printed it using the correct %p-format - corresponds to the hex-value of the address (yet in decimal format).
%d prints an int, in decimal.
%p prints a pointer (strictly speaking, a void * pointer) in an implementation-defined way, typically in hexadecimal.
On a machine where ints and pointers have different sizes, trying to print a pointer using %d will typically give a meaningless result. For example, on a lot of machines these days, ints are 32 bits while pointers are 64 bits. So if you try to print a pointer using %d, what you might get is half the pointer value, in decimal.
(Strictly speaking, trying to print a pointer using %d is undefined, no matter what the relative sizes are.)
Bottom line, use the right printf specifier for the job. Use %d to print ints. Use %p to print pointers. Don't try to mix 'n' match.
I tried to print character as a float in printf and got output 0. What is the reason for this.
Also:
char c='z';
printf("%f %X",c,c);
is giving some weird output for hexadecimal while output is correct when I do this:
printf("%X",c);
why is it so?
The printf() function is a variadic function, which means that you can pass a variable number of arguments of unspecified types to it. This also means that the compiler doesn't know what type of arguments the function expects, and so it cannot convert the arguments to the correct types. (Modern compilers can warn you if you get the arguments wrong to printf, if you invoke it with enough warning flags.)
For historical reasons, you can not pass an integer argument of smaller rank than int, or a floating type of smaller rank than double to a variadic function. A float will be converted to double and a char will be converted to int (or unsigned int on bizarre implementations) through a process called the default argument promotions.
When printf parses its parameters (arguments are passed to a function, parameters are what the function receives), it retrieves them using whatever method is appropriate for the type specified by the format string. The "%f" specifier expects a double. The "%X" specifier expects an unsigned int.
If you pass an int and printf tries to retrieve a double, you invoke undefined behaviour.
If you pass an int and printf tries to retrieve an unsigned int, you invoke undefined behaviour.
Undefined behaviour may include (but is not limited to) printing strange values, crashing your program or (the most insidious of them all) doing exactly what you expect.
Source: n1570 (The final public draft of the current C standard)
You need to use a cast operator like this:
char c = 'z';
printf("%f %X", (float)c, c);
or
printf("%f %X", (double)c, c);
In Xcode, if I do not do this, I get the warning:
Format specifies specifies 'double' but the argument has type 'char', and the output is 0.000000.
I tried to print character as a float in printf and got output 0. What is the reason for this.
The question is, what value did you expect to see? Why would you expect something other than 0?
The short answer to your question is that the behavior of printf is undefined if the type of the argument doesn't match the conversion specifier. The %f conversion specifier expects its corresponding argument to have type double; if it isn't, all bets are off, and the exact output will vary.
To understand the floating point issue, consider reading: http://en.wikipedia.org/wiki/IEEE_floating_point
As for hexadecimal, let me guess.. the output was something like... 99?
This is because of encodings.. the machine has to represent information in some format, and usually that format entails either giving meanings to certain bits in a number, or having a table of symbols to numbers, or both
Floating points are sometimes represented as a (sign,mantissa,exponent) triplet all packed in a 32 or 64 bit number - characters are sometimes represented in a format named ASCII, which establishes which number corresponds to each character you type
Because printf, like any function that work with varargs, eg: int foobar(const char fmt, ...) {} tries to interpret its parameter to certain type.
If you say "%f", then pass c (as a char), then printf will try to read a float.
You can read more here: var_arg (even if this is C++, it still applies).
I tried to print character as a float in printf and got output 0. What is the reason for this.
Also:
char c='z';
printf("%f %X",c,c);
is giving some weird output for hexadecimal while output is correct when I do this:
printf("%X",c);
why is it so?
The printf() function is a variadic function, which means that you can pass a variable number of arguments of unspecified types to it. This also means that the compiler doesn't know what type of arguments the function expects, and so it cannot convert the arguments to the correct types. (Modern compilers can warn you if you get the arguments wrong to printf, if you invoke it with enough warning flags.)
For historical reasons, you can not pass an integer argument of smaller rank than int, or a floating type of smaller rank than double to a variadic function. A float will be converted to double and a char will be converted to int (or unsigned int on bizarre implementations) through a process called the default argument promotions.
When printf parses its parameters (arguments are passed to a function, parameters are what the function receives), it retrieves them using whatever method is appropriate for the type specified by the format string. The "%f" specifier expects a double. The "%X" specifier expects an unsigned int.
If you pass an int and printf tries to retrieve a double, you invoke undefined behaviour.
If you pass an int and printf tries to retrieve an unsigned int, you invoke undefined behaviour.
Undefined behaviour may include (but is not limited to) printing strange values, crashing your program or (the most insidious of them all) doing exactly what you expect.
Source: n1570 (The final public draft of the current C standard)
You need to use a cast operator like this:
char c = 'z';
printf("%f %X", (float)c, c);
or
printf("%f %X", (double)c, c);
In Xcode, if I do not do this, I get the warning:
Format specifies specifies 'double' but the argument has type 'char', and the output is 0.000000.
I tried to print character as a float in printf and got output 0. What is the reason for this.
The question is, what value did you expect to see? Why would you expect something other than 0?
The short answer to your question is that the behavior of printf is undefined if the type of the argument doesn't match the conversion specifier. The %f conversion specifier expects its corresponding argument to have type double; if it isn't, all bets are off, and the exact output will vary.
To understand the floating point issue, consider reading: http://en.wikipedia.org/wiki/IEEE_floating_point
As for hexadecimal, let me guess.. the output was something like... 99?
This is because of encodings.. the machine has to represent information in some format, and usually that format entails either giving meanings to certain bits in a number, or having a table of symbols to numbers, or both
Floating points are sometimes represented as a (sign,mantissa,exponent) triplet all packed in a 32 or 64 bit number - characters are sometimes represented in a format named ASCII, which establishes which number corresponds to each character you type
Because printf, like any function that work with varargs, eg: int foobar(const char fmt, ...) {} tries to interpret its parameter to certain type.
If you say "%f", then pass c (as a char), then printf will try to read a float.
You can read more here: var_arg (even if this is C++, it still applies).
While using printf with %d as format specifier and giving a float as an argument e.g, 2.345, it prints 1546188227. So, I understand that it may be due to the conversion of single point float precision format to simple decimal format. But when we print 2.000 with the %d as format specifier, then why it prints 0 only ?
Please help.
Format specifier %d can only be used with values of type int (and compatible types). Trying to use %d with float or any other types produces undefined behavior. That's the only explanation that truly applies here. From the language point of view the output you see is essentially random. And you are not guaranteed to get any output at all.
If you are still interested in investigating the specific reason for the output you see (however little sense it makes), you'll have to perform a platform-specific investigation, because the actual behavior depends critically on various implementation details. And you are not even mentioning your platform in your post.
In any case, as a side note, note that it is impossible to pass float values as variadic arguments to variadic functions. float values in such cases are always converted to double and passed as double. So in your case it is double values you are attempting to print. Behavior is still undefined though.
Go here, enter 2.345, click "rounded". Observe 64-bit hex value: 4002C28F5C28F5C3. Observe that 1546188227 is 0x5c28f5c3.
Now repeat for 2.00. Observe that 64-bit hex value is 4000000000000000
P.S. When you say that you give a float argument, what you apparently mean is that you give a double argument.
Here what ISO/IEC 9899:1999 standard $7.19.6 states:
If a conversion specification is invalid, the behavior is undefined.239)
If any argument is not the correct type for the corresponding conversion
specification, the behavior is undefined.
If you're trying to make it print integer values, cast the floats to ints in the call:
printf("ints: %d %d", (int) 2.345, (int) 2.000);
Otherwise, use the floating point format identifier:
printf("floats: %f %f", 2.345, 2.000);
When you use printf with wrong format specifier for the corresponding argument, the result is undefined behavior. It could be anything and may differ from one implementation to another. Only correct use of format specified has defined behavior.
First a small nitpick: The literal 2.345 is actually of the type double, not float, and besides, even a float, such as the literal 2.345f, would be converted to double when used as an argument to a function that takes a variable number of arguments, such as printf.
But what happens here is that the (typically) 64 bits of the double value is sent to printf, and then it interprets (typically) 32 of those bits as an integer value. So it just happens that those bits were zero.
According to the standard, this is what is called undefined behavior: The compiler is allowed to do anything at all.