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void sum() { int sum; // <-- one `sum` for (int i=0,sum=0;i<=10;sum+=i,++i) {} // <-- second `sum` cout << sum << endl; } Only a declaration statement like that can be in the first clause of the for preamble (think about whether sum=0,int i=0 would be valid elsewhere in your code), but you can workaround this issue by pulling out the ...
Question 5 int sum =0; for(i=0; i< 10; i++); sum += 1; What is the value of sum after the above code 63 11 Question 7 int i = 10; int sum = 0; do { sum = 25; } until (i>10): What happens when this code executes? infinite loop sum becomes 25, the loop executes once sum remains zero, the loop never execute Question 8 int i; int sum; Sum = 0; for ...
Consider the following C program: C #include int main() { int a[] = {2, 4, 6, 8, 10}; int i, sum = 0, *b = a + 4; for (i = 0; i < 5; i++ ) sum = sum + (*b ...
for(i = 0; i < 10; i++) is the header of a for loop. A for loop is a short hand notation for a particular type of while loops, but can also be used in different ways. A for loop, but also a while loop, consists of two parts: a header (for(...) or while(...)) and the body ({...}In the header the control logic for the loop is set up, the body is where the actual code is found.
Forget about the ++ for the moment, they're just a distraction.. int a = 5; int i = 0; int x = (i, a); Sets the value of x to 5. The i is evaluated and discarded, then the a is evaluated and assigned to x.. In your loop, the post-increment a++ does what it always does; returns the current value and then increments the variable. So x takes the value of a before it is incremented, then a's value ...
Analyze the following fragment: double sum = 0; double d = 0; while (d != 10.0) {d += 0.1; sum += sum + d;} A. The program does not compile because sum and d are declared double, but assigned with integer value 0. B. The program never stops because d is always 0.1 inside the loop. C. The program may not stop because of the phenomenon referred ...
Good notes 1)aim: find sum and reverse of number using algorithm: step step step step step step step step step step step start read number num set and repeat
In the second example you have for (int i = 0; i < 10; i=i+2) This means the loop value i will start at 0, then go 2, 4, 6 etc. so you will be adding a[0], a[2], a[4] etc. all the way up to a[8] In the more complex examples you have a different starting value for your loop...
However if you do, first define it and then use it. Also the return type of change function should be void as this function is not returning anything and if you don't specify the return type the default is int. Change change ( int *b, int n ) to void change ( int *b, int n ).
In every round, the variable i is getting incremented before a new loop round (i++ in the loop header) and after the line sum += i++;.. This leads to i being 0, 2, 4 consecutively for each time, the line mentioned is called. After i=4 and i being incremented by the loop, the loop stops. Thus, sum=0+2+4=6 is your output. This behavior is to be expected because of the postfix incrementation ...