Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

What's wrong with this?

Computer Science
See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

1 Attachment
SALARIO is an array but doesn't have any values so how can anything be printed from it . What is this C?
but apart from the values? is it all ok? i think it's C++ my professor told me to run it in Dev C++

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

well can float be an array?
thats in java
not sure about FLOAT though
well, if you're no sure, i dont know what i'm going to do :P
do you know what an API is?
google C++ API array float
yes, i do :)
great then look up the stuff I said you need to make an array with values....see what datatypes you can use FLOAT is a data type
allot of languages are using strong data typing like this.....var testVar0:String ="testVariable"; thats AS 3.0
yeah, but i'll look it up and try again. Thank you :))
you have a structural issue more than a API issue.....seperate variable .....methods .....functions in your head. create variable>>>>method does something to the variable>>>>>method ouputs variable and the new variable is stored either in the same variable or a newVariable Then there are events that can call functions from within other functions....not in C++ though that I'm aware.
what you have is either a C or C++ program
when you say whats wrong with this - what do you mean exactly?
FFS C and C++ are the same language its just C++ has a whole bunch of prebuilt methods made for it. Its a higher level language generation 3 I think.
correct - which is why it could be either
nickymarden: what is it that is making you say there is something wrong with it?
you don't see the problem?
I don't see the problem
I suppose this line populates salario array scanf("%f",&salario[i]);
I don't know C mush though
I guess there is no problem and they don't know how to compile
oh well have fun you guys I'll be back later too many cooks in the kitchen
i don't see a problem, but it isn't running, so i assumed there was one :)
What isn't wrong with it may be a better question.. I've patched it up into a running state, but I'll let you diff and figure out what changed.. :) #include int main(void){ float salario[50]; for(int i=0; i<50; i++){ printf("DIGITE O SALARIO\n"); scanf("%f",&salario[i]); } float aumento = 0; printf("QUAL FOI O AUMENTO?\n"); scanf("%f", &aumento); for(int i=0; i<50; i++){ printf("O SALARIO ANTES ERA %f\n", salario[i]); salario[i]=salario[i] + salario[i]*(aumento/100); printf ("O SALARIO AGORA E %f\n", salario[i]); } return(0); }
Yep, there's problems. On the first for loop, for(int i=0; i<50; l++), you used an L instead of 'i' for some reason. Replace it with for(int i=0; i<50; i++). Secondly, on scanf("%", %aumento), you have two things wrong. 1, the first parameter must include a character format. Since you want to read a float, use %f. Second, the second parameter is an address to the float it is read into. The address of operator is &. So, &aumento. So it's scanf("%f", &aumento); This is most likely C++. It is possibly C99.
Thaaank you :)) Helped a lot, it's running now :P and yeah, i asked my professor and it's C++ after all :)
well ur prof wrong its c, 1. printf,scanf is used in c, in c++ we use cout and cin 2. its better to declare ur variable outside a loop, u declared 'i' twice in ur code 3. in ur first for loop use for(i=0; i<50; 1++), that should be for(i=0;i<50; i++) 4. and use a system("pause"); before the return 0 function, its good practice. it will pause the program and wait for user input
yeah, i got it to run, thank you :))
@kevissen - 1. Just because it uses printf does not mean it's C. 2. You should not declare a variable used only for looping outside of the for loop 3. I pointed that out, but I agree. 4. This is easily the most offending thing and I am wondering how well you know C or C++. system("pause") is a HORRIBLE practice, and isn't in the C/C++ spec. If you want to wait for user input, just insert a scanf or cin.

Not the answer you are looking for?

Search for more explanations.

Ask your own question