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#include <stdio.h>
int sum(int a);
int main()
{
int a;
printf("Enter a value: ");
scanf("%d", &a);
printf("%d", sum(a));
return 0;
}
int sum(int a)
{
if (a == 1)
{
return 1;
}
return a + sum(a - 1);
}
When the input is 5 the output is 15 (which is right),
but when the return is, return a + sum(--a);
(for the same input 5) the output is 11
The behaviour of a + sum(--a) is undefined. The compiler has a lot of freedom as to when and how often it reads a, and when and how often it modifies a. Avoid both reading and modifying the same variable in an expression.
When you wrote
a + sum(a - 1)
, a rough translation of that C code into English is:
"Take a plus the result of the sum function applied to the quantity a-1."
That makes sense, and it's just what you want.
If, on the other hand, you write
a + sum(--a)
, now what you're saying is,
"Take a plus the result of the sum function applied to, hang on a minute, I want to take the variable a minus 1 and assign it back to a and then, where was I, pass it to sum()."
Is that really what you want to do? Why would you want to modify a's value in the middle of calling sum()? You wouldn't want to do that, you don't need to do that, that has nothing to do with the problem of calling sum(a - 1). So don't write it that way. It's confusing to you and me, and what's worse, it's so confusing that the compiler can't figure it out, either. It's so confusing that it's undefined behavior, meaning that the compiler isn't even required to figure out what you mean, and the compiler is explicitly allowed to fall back and punt, and compile your program into something that doesn't work — which is precisely what you discovered when you tried it.
See the canonical SO question Why are these constructs using pre and post-increment undefined behavior? for much (much!) more on undefined expressions like these.
The following code is here with keep the C Language Syntax:
#include <stdio.h>
int func(int a, int b){
if (b==0)
return 0;
else return func(a,b);
}
int main(){
printf("%d \n", func(func(1,1),func(0,0)));
return 0;
}
What is the output of this code at 1) run with standard C, 2) with any
language that has call by need property, Then:
in (1) the programs loop into infinite call and in (2) we have ouptut zero !! this is an example solved by TA in programming language course, any idea to
describe it for me? thanks
1) In C (which uses strict evaluation semantics) we get infinite recursion because in strict evaluation arguments are evaluated before a function is called. So in f(f(1,1), f(0,0)) f(1,1) and f(0,0) are evaluated before the outer f (which one of the two arguments is evaluated first is unspecified in C, but that does not matter). And since f(1,1) causes infinite recursion, we get infinite recursion.
2) In a language using non-strict evaluation (be it call-by-name or call-by-need) arguments are substituted into the function body unevaluated and are only evaluated when and if they're needed. So the outer call to f is evaluated first as such:
if (f(0, 0) == 0)
return 0;
else return f(f(1,1), f(0,0));
So when evaluating the if, we need to evaluate f(0,0), which simply evaluates to 0. So we go into the then-branch of the if and never execute the else-branch. Since all calls to f are only used in the else-branch, they're never needed and thus never evaluated. So there's no recursion, infinite or otherwise, and we just get 0.
With C, in general, it is not defined the order of arguments a and b evaluation with a function like int func(int a, int b)
Obviously evaluating func(1,1) is problematic and the code suffers from that regardless if func(1,1) is evaluated before/after/simultaneous with func(0,0)
Analysis of func(a,b) based on need may conclude that if b==0, no need to call func() and then replace with 0.
printf("%d \n", func(func(1,1),func(0,0)));
// functionally then becomes
printf("%d \n", func(func(1,1),0));
Applied again and
// functionally then becomes
printf("%d \n", 0);
Of course this conclusion is not certain as the analysis of b != 0 and else return func(a,b); leads to infinite recursion. Such code may have a useful desired side-effect (e.g. stack-overflow and system reset.) So the analysis may need to be conservative and not assume func(1,1) will ever return and not optimize out the call even if it optimized out the func(0,0) call.
To address the first part,
The C Draft, 6.5.2.2->10(Function calls) says
The order of evaluation of ... the actual arguments... is unspecified.
and for such reason, something such as
printf("%d%d",i++,++i);
has undefined behaviour, because
both ++i and i++ has side-effects, ie incrementing the value of i by one.
The comma inside printf is just a separator and NOT a [ sequence point ].
Even though function call itself is a sequence point, for the above reason, the order in which two modifications of i take place is not defined.
In your case
func(func(1,1),func(0,0))
though, the arguments for outer func ie func(1,1) or func(0,0) have no bearing on each other contrary to the case shown above. Any order of evaluation of these arguments eventually leads to infinite recursion and so the program crashes due to depleted memory.
how is your day :),
Take a look at the below program, the program written below is to calculate the sum of first n natural numbers, the problem is that i get the sum of n-1 natural numbers, can anybody explain why ?
and can anybody also explain why a-- instead of --a.
#include<stdio.h>
main()
{
int a,sum;
printf("Enter a number.");
scanf("%d",&a);
sum=sumnat(a);
printf("Sum of the first %d natural numbers is %d.",a,sum);
}
sumnat(a)
{
int b;
if(a==0)
{
return 0;
}
else
{
b=a+sumnat(--a);
return(b);
}
}
There were several errors, the greatest of which was undefined behaviour in the expression which uses a and also a modified value of a. You should also define your function properly, not rely on default values provided by the compiler.
#include <stdio.h>
int sumnat(int a); // function prototype
int main(void) // correct signature
{
int a, sum;
printf("Enter a number. ");
scanf("%d", &a);
sum = sumnat(a);
printf("Sum of the first %d natural numbers is %d.", a, sum);
return 0;
}
int sumnat(int a) // function has a return type and argument type
{
if(a == 0)
{
return 0;
}
return a + sumnat(a - 1); // there was no need to decrement `a`
}
Program session
Enter a number. 5
Sum of the first 5 natural numbers is 15.
Your program works for me, using gcc on Mac OSX. However, it will not work everywhere, because of this line:
b=a+sumnat(--a);
--a decrements a, but if it does so before the addition, then your result will be wrong. I'm not sure C is required to evaluate expressions strictly left-to-right (I don't think it is). At any rate, since you don't use a after that line, you could fix things this way:
b=a+sumnat(a-1);
As #self says, you should fix the program to handle negative values, and it would be a good idea to consider what is the largest natural number whose sum you can compute this way (and why that is).
There is a difference between them. One first subracts from a and than goes in the function while the other frist goes in... so it never gets subracted and you go to inifinit stack.
"and can anybody also explain why a-- instead of --a"
When you use the prefix operator --a the decrease is done before anything else, while the postfix operator a-- happens after the rest of the expression is resolved, so, lets say, while debugging your code, in a particular moment, a = 5
since the line
b=a+sumnat(--a);
is using the prefix version of the operator, the decrement would happen immediately, making a=4 and then the function sumnat would be called with argument 4
b=a+sumnat(a--);
in this case the postfix operator is being used, so first the function sumnat would be called with the argument 5, since that is the value of a in that moment, then, only when the function returns a value (which would never happen in your example, since it would be called multiple times with the same value, never reaching 0) the decrement would happen
I'm just learning some C, or rather, getting a sense of some of the arcane details. And I was using VTC advanced C programming in which I found that the sequence points are :
Semicolon
Comma
Logical OR / AND
Ternary operator
Function calls (Any expression used as an argument to a function call is finalized the call is made)
are all these correct ?. Regarding the last one I tried:
void foo (int bar) { printf("I received %d\n", bar); }
int main(void)
{
int i = 0;
foo(i++);
return 0;
}
And it didnt print 1, which according to what the VTC's guy said and if I undertood correctly, it should, right ?. Also, are these parens in the function call the same as the grouping parens ? (I mean, their precedence). Maybe it is because parens have higher precedence than ++ but I've also tried foo((i++)); and got the same result. Only doing foo(i = i + 1); yielded 1.
Thank you in advance. Please consider that I'm from South America so if I wasnt clear or nothing makes sense please, oh please, tell me.
Warmest regards,
Sebastian.
Your code is working like it should and it has nothing to do with sequence points. The point is that the postfix ++ operator returns the non-incremented value (but increments the variable by 1 nonetheless).
Basically:
i++ – Increment i by one and return the previous value
++i – Increment i by one and return the value after the increment
The position of the operator gives a slight hint for its semantics.
Sequence means i++ is evaluted before foo is invoked.
Consider this case (I am not printing bar!):
int i = 0;
void foo (int bar) { printf("i = %d\n", i); }
int main(void){
foo(i++);
return 0;
}
i = 1 must be printed.
C implements pass-by-value semantics. First i ++ is evaluated, and the value is kept, then i is modified (this may happen any time between the evaluation and the next sequence point), then the function is entered with the backup value as the argument.
The value passed into a function is always the same as the one you would see if using the argument expression in any other way. Other behavior would be fairly surprising, and make it difficult to refactor common subexpressions into functions.
When you do something like:
int i = 0, j;
j = i++;
the value of i is used first and then incremented. hence in your case the values of i which is 0 is used (hence passed to your function foo) and then incremented. the incremented values of i (now 1) will be available only for main as it is its local variable.
If you want to print 1 the do call foo this way:
foo(++i);
this will print 1. Rest you know, why!
How to get and evaluate the Expressions from a string in C
char *str = "2*8-5+6";
This should give the result as 17 after evaluation.
Try by yourself. you can Use stack data structure to evaluate this string here is reference to implement (its in c++)
stack data structre for string calcualtion
You have to do it yourself, C does not provide any way to do this. C is a very low level language. Simplest way to do it would be to find a library that does it, or if that does not exist use lex + yacc to create your own interpreter.
A quick google suggests the following:
http://www.gnu.org/software/libmatheval/
http://expreval.sourceforge.net/
You should try TinyExpr. It's a single C source code file (with no dependencies) that you can add to your project.
Using it to solve your problem is just:
#include <stdio.h>
#include "tinyexpr.h"
int main()
{
double result = te_interp("2*8-5+6", 0);
printf("Result: %f\n", result);
return 0;
}
That will print out: Result: 17
C does not have a standard eval() function.
There are lots of libraries and other tools out there that can do this.
But if you'd like to learn how to write an expression evaluator yourself, it can be surprisingly easy. It is not trivial: it is actually a pretty deeply theoretical problem, because you're basically writing a miniature parser, perhaps built on a miniature lexical analyzer, just like a real compiler.
One straightforward way of writing a parser involves a technique called recursive descent. Writing a recursive descent parser has a lot in common with another great technique for solving big or hard problems, namely by breaking the big, hard problem up into smaller and hopefully easier subproblems.
So let's see what we can come up with. We're going to write a function int eval(const char * expr) that takes a string containing an expression, and returns the int result of evaluating it. But first let's write a tiny main program to test it with. We'll read a line of text typed by the user using fgets, pass it to our expr() function, and print the result.
#include <stdio.h>
int eval(const char *expr);
int main()
{
char line[100];
while(1) {
printf("Expression? ");
if(fgets(line, sizeof line, stdin) == NULL) break;
printf(" -> %d\n", eval(line));
}
}
So now we start writing eval(). The first question is, how will we keep track of how how far we've read through the string as we parse it? A simple (although mildly cryptic) way of doing this will be to pass around a pointer to a pointer to the next character. That way any function can move forwards (or occasionally backwards) through the string. So our eval() function is going to do almost nothing, except take the address of the pointer to the string to be parsed, resulting in the char ** we just decided we need, and calling a function evalexpr() to do the work. (But don't worry, I'm not procrastinating; in just a second we'll start doing something interesting.)
int evalexpr(const char **);
int eval(const char *expr)
{
return evalexpr(&expr);
}
So now it's time to write evalexpr(), which is going to start doing some actual work. Its job is to do the first, top-level parse of the expression. It's going to look for a series of "terms" being added or subtracted. So it wants to get one or more subexpressions, with + or - operators between them. That is, it's going to handle expressions like
1 + 2
or
1 + 2 - 3
or
1 + 2 - 3 + 4
Or it can read a single expression like
1
Or any of the terms being added or subtracted can be a more-complicated subexpression, so it will also be able to (indirectly) handle things like
2*3 + 4*5 - 9/3
But the bottom line is that it wants to take an expression, then maybe a + or - followed by another subexpression, then maybe a + or - followed by another subexpression, and so on, as long as it keeps seeing a + or -. Here is the code. Since it's adding up the additive "terms" of the expression, it gets subexpressions by calling a function evalterm(). It also needs to look for the + and - operators, and it does this by calling a function gettok(). Sometimes it will see an operator other than + or -, but those are not its job to handle, so if it sees one of those it "ungets" it, and returns, because it's done. All of these functions pass the pointer-to-pointer p around, because as I said earlier, that's how all of these functions keep track of how they're moving through the string as they parse it.
int evalterm(const char **);
int gettok(const char **, int *);
void ungettok(int, const char **);
int evalexpr(const char **p)
{
int r = evalterm(p);
while(1) {
int op = gettok(p, NULL);
switch(op) {
case '+': r += evalterm(p); break;
case '-': r -= evalterm(p); break;
default: ungettok(op, p); return r;
}
}
}
This is some pretty dense code, Stare at it carefully and convince yourself that it's doing what I described. It calls evalterm() once, to get the first subexpression, and assigns the result to the local variable r. Then it enters a potentially infinite loop, because it can handle an arbitrary number of added or subtracted terms. Inside the loop, it gets the next operator in the expression, and uses it to decide what to do. (Don't worry about the second, NULL argument to gettok; we'll get to that in a minute.)
If it sees a +, it gets another subexpression (another term) and adds it to the running sum. Similarly, if it sees a -, it gets another term and subtracts it from the running sum. If it gets anything else, this means it's done, so it "ungets" the operator it doesn't want to deal with, and returns the running sum -- which is literally the value it has evaluated. (The return statement also breaks the "infinite" loop.)
At this point you're probably feeling a mixture of "Okay, this is starting to make sense" but also "Wait a minute, you're playing pretty fast and loose here, this is never going to work, is it?" But it is going to work, as we'll see.
The next function we need is the one that collects the "terms" or subexpressions to be added (and subtracted) together by evalexpr(). That function is evalterm(), and it ends up being very similar -- very similar -- to evalexpr. Its job is to collect a series of one or more subexpressions joined by * and/or /, and multiply and divide them. At this point, we're going to call those subexpressions "primaries". Here is the code:
int evalpri(const char **);
int evalterm(const char **p)
{
int r = evalpri(p);
while(1) {
int op = gettok(p, NULL);
switch(op) {
case '*': r *= evalpri(p); break;
case '/': r /= evalpri(p); break;
default: ungettok(op, p); return r;
}
}
}
There's actually nothing more to say here, because the structure of evalterm ends up being exactly like evalexpr, except that it does things with * and /, and it calls evalpri to get/evaluate its subexpressions.
So now let's look at evalpri. Its job is to evaluate the three lowest-level, but highest-precedence elements of an expression: actual numbers, and parenthesized subexpressions, and the unary - operator.
int evalpri(const char **p)
{
int v;
int op = gettok(p, &v);
switch(op) {
case '1': return v;
case '-': return -evalpri(p);
case '(':
v = evalexpr(p);
op = gettok(p, NULL);
if(op != ')') {
fprintf(stderr, "missing ')'\n");
ungettok(op, p);
}
return v;
}
}
The first thing to do is call the same gettok function we used in evalexpr and evalterm. But now it's time to say a little more about it. It is actually the lexical analyzer used by our little parser. A lexical analyzer returns primitive "tokens". Tokens are the basic syntactic elements of a programming language. Tokens can be single characters, like + or -, or they can also be multi-character entities. An integer constant like 123 is considered a single token. In C, other tokens are keywords like while, and identifiers like printf, and multi-character operators like <= and ++. (Our little expression evaluator doesn't have any of those, though.)
So gettok has to return two things. First it has to return a code for what kind of token it found. For single-character tokens like + and - we're going to say that the code is just the character. For numeric constants (that is, any numeric constant), we're going to say that gettok is going to return the character 1. But we're going to need some way of knowing what the value of the numeric constant was, and that, as you may have guessed, is what the second, pointer argument to the gettok function is for. When gettok returns 1 to indicate a numeric constant, and if the caller passes a pointer to an int value, gettok will fill in the integer value there. (We'll see the definition of the gettok function in a moment.)
At any rate, with that explanation of gettok out of the way, we can understand evalpri. It gets one token, passing the address of a local variable v in which the value of the token can be returned, if necessary. If the token is a 1 indicating an integer constant, we simply return the value of that integer constant. If the token is a -, this is a unary minus sign, so we get another subexpression, negate it, and return it. Finally, if the token is a (, we get another whole expression, and return its value, checking to make sure that there's another ) token after it. And, as you may notice, inside the parentheses we make a recursive call back to the top-level evalexpr function to get the subexpression, because obviously we want to allow any subexpression, even one containing lower-precedence operators like + and -, inside the parentheses.
And we're almost done. Next we can look at gettok. As I mentioned, gettok is the lexical analyzer, inspecting individual characters and constructing full tokens from them. We're now, finally, starting to see how the passed-around pointer-to-pointer p is used.
#include <stdlib.h>
#include <ctype.h>
void skipwhite(const char **);
int gettok(const char **p, int *vp)
{
skipwhite(p);
char c = **p;
if(isdigit(c)) {
char *p2;
int v = strtoul(*p, &p2, 0);
*p = p2;
if(vp) *vp = v;
return '1';
}
(*p)++;
return c;
}
Expressions can contain arbitrary whitespace, which is ignored, so we skip over that with an auxiliary function skipwhite. And now we look at the next character. p is a pointer to pointer to that character, so the character itself is **p. If it's a digit, we call strtoul to convert it. strtoul helpfully returns a pointer to the character following the number it scans, so we use that to update p. We fill in the passed pointer vp with the actual value strtoul computed for us, and we return the code 1 indicating an integer constant.
Otherwise -- if the next character isn't a digit -- it's an ordinary character, presumably an operator like + or - or punctuation like ( ), so we simply return the character, after incrementing *p to record the fact that we've consumed it. Properly "incrementing" p is mildly tricky: it's a pointer to a pointer, and we want to increment the pointed-to pointer. If we wrote p++ or *p++ it would increment the pointer p, so we need (*p)++ to say that it's the pointed-to pointer that we want to increment. (See also C FAQ 4.3.)
Two more utility functions, and then we're done. Here's skipwhite:
void skipwhite(const char **p)
{
while(isspace(**p))
(*p)++;
}
This simply skips over zero or more whitespace characters, as determined by the isspace function from <ctype.h>. (Again, we're taking care to remember that p is a pointer-to-pointer.)
Finally, we come to ungettok. It's a hallmark of a recursive descent parser (or, indeed, almost any parser) that it has to "look ahead" in the input, making a decision based on the next token. Sometimes, however, it decides that it's not ready to deal with the next token after all, so it wants to leave it there on the input for some other part of the parser to deal with later.
Stuffing input "back on the input stream", so to speak, can be tricky. This implementation of ungettok is simple, but it's decidedly imperfect:
void ungettok(int op, const char **p)
{
(*p)--;
}
It doesn't even look at the token it's been asked to put back; it just backs the pointer up by 1. This will work if (but only if) the token it's being asked to unget is in fact the most recent token that was gotten, and if it's not an integer constant token. In fact, for the program as written, and as long as the expression it's parsing is well-formed, this will always be the case. It would be possible to write a more complicated version of gettok that explicitly checked for these assumptions, and that would be able to back up over multi-character tokens (such as integer constants) if necessary, but this post has gotten much longer than I had intended, so I'm not going to worry about that for now.
But if you're still with me, we're done! And if you haven't already, I encourage you to copy all the code I've presented into your friendly neighborhood C compiler, and try it out. You'll find, for example, that 1 + 2 * 3 gives 7 (not 9), because the parser "knows" that * and / have higher precedence than + and -. Just like in a real compiler, you can override the default precedence using parentheses: (1 + 2) * 3 gives 9. Left-to-right associativity works, too: 1 - 2 - 3 is -4, not +2. It handles plenty of complicated, and perhaps surprising (but legal) cases, too: (((((5))))) evaluates to just 5, and ----4 evaluates to just 4 (it's parsed as "negative negative negative negative four", since our simplified parser doesn't have C's -- operator).
This parser does have a pretty big limitation, however: its error handling is terrible. It will handle legal expressions, but for illegal expressions, it will either do something bizarre, or just ignore the problem. For example, it simply ignores any trailing garbage it doesn't recognize or wasn't expecting -- the expressions 4 + 5 x, 4 + 5 %, and 4 + 5 ) all evaluate to 9.
Despite being somewhat of a "toy", this is also a very real parser, and as we've seen it can parse a lot of real expressions. You can learn a lot about how expressions are parsed (and about how compilers can be written) by studying this code. (One footnote: recursive descent is not the only way to write a parser, and in fact real compilers will usually use considerably more sophisticated techniques.)
You might even want to try extending this code, to handle other operators or other "primaries" (such as settable variables). Once upon a time, in fact, I started with something like this and extended it all the way into an actual C interpreter.