int main(int argc, char ** argv) { int i = 0; i = i++ + ++i; printf("%d\n", i); // 3 i = 1; i = (i++); printf("%d\n", i); // 2 Should be 1, no ? volatile int u = 0; u = u++ + ++u; printf("%d\n", u); // 1 u = 1; u = (u++); printf("%d\n", u); // 2 Should also be one, no ? register int v = 0; v...
x = 1 i expect the answer to be 11, but it comes out to be 12. ...
Our class was asked this question by the C programming prof: You are given the code: int x=1; printf("%d",++x,x+1); What output will it always produce ? Most students said undefined behavior. Can anyone help me understand why it is so? Thanks for the edit and the answers but I'm still confused. ...
Possible Duplicate: Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc) int main() { int a=5,s; s=++a + ++a; printf("%d",a); printf("%d",s); } output is 7 and 14 BUT int main() { int a, s; printf("Enter value of a"); scanf ("%d",&a); s=++a + ++a; printf("%d",a); printf("%d",s); } input user give...
Hello Once again, our best loved "i=i--" -like issues. In C99 we have: 6.5 Expressions #2: Between the previous and next sequence point an object shall have its stored value modified at most once 70) This paragraph renders !!undefined!! statement expressions such as i = ++i + 1; But for undefinded behavior there ca...
Hi, after reading about sequence points, I learned that i = ++i is undefined. So how about this code: int i; int *p = &i; int *q = &i; *p = ++(*q); // that should also be undefined right? Let's say if initialization of p and q depends on some (complicated) condition. And they may be pointing to same object like in above ca...
int main() { int i=3; (++i)++; printf("%d",i); } This programs works with g++ compiler but not gcc. If i write i++++ or ++i++ it doesn't work in cpp also. I think there is difference between c-expression and c++-expression. Can somebody explain about L-value and R-value ? ...
#include<stdio.h> int main() { int i=7,j; j=(i++,++i,j*i); return 0; } j=(i++,++i,j*i);Is this well defined ? Let me clear my doubt. ...
Possible Duplicate: C programming: is this undefined behavior? #include<stdio.h> main() { int i=5; printf("%d%d%d",i,i++,++i); } my expected output is 556. But when i executed it the result is 767. how? ...
int val = 5; printf("%d",++val++); //gives compilation error : '++' needs l-value int *p = &val; printf("%d",++*p++); //no error Could someone explain these 2 cases? Thanks. ...
Why does the following compile in C++? int phew = 53; ++++++++++phew ; The same code fails in C, why? ...
Why ++x || ++y && ++z calculate ++x firstly? However,Operator && is higher than ||? ...
I recently came along this method for swapping the values of two variables without using a third variable. a^=b^=a^=b But when I tried the above code on different compilers, I got different results, some gave correct results, some didn't. Is anything terribly wrong with the code? ...
Will the right side of the expression get evaluated first or the left ? void main () { int i = 0 , a[3] ; a[i] = i++; printf ("%d",a[i]) ; } ...
The C-faq says that the code: int i = 7; printf("%d\n", i++ * i++); prints 49. Regardless of the order of evaluation, shouldn't it print 56? When I ran this code on my Turbo C 3.0 compiler it gave me the output of 56. Why is there a contradiction? ...
Here is my function: void abc(char *def, unsigned int w, unsigned int x, unsigned int y, unsigned int z) { printf("val 1 : %d\n", w); printf("val 2 : %d\n", x); printf("val 3 : %d\n", y); printf("val 4 : %d\n", z); } and here is where I call this function: unsigned int exp[4] = { 1, 2, 3, 4 }; unsigned short count = 0; a...
Possible Duplicate: How do we explain the result of the expression (++x)+(++x)+(++x)? int i=2; i = ++i + ++i + ++i; Which is more correct? Java's result of 12 or C = 13. Or if not a matter of correctness, please elaborate. Thank you. ...
Possible Duplicate: Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc) What is the difference between i = ++i; and ++i; where i is an integer with value 10? According to me both do the same job of incrementing i i.e after completion of both the expressions i =11. ...
Possible Duplicate: Could anyone explain these undefined behaviors (i = i++ + ++i , i = i++, etc) I am using Visual C++ 6.0 and facing trouble predicting the output of the following program: int main(int argc,char **argv) { int a; int b; a=1; b = (++a) + (a++) + (a++); printf("%d",b); return 0; } ...
We've been taught various syntax and told how to write definitions, but we've never written any code an ran it. What is the order that Scheme code runs in? Thanks! ...