What does double semicolon mean in c? [closed] - c

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For example:
void thisIsAnExample(Hello* bye, char* name, int num, in* arr, int* sum){
GoodBye x;;
x.funName = name;
.
.
.

It doesn't mean anything. It's just an extra semicolon. You can delete it (leaving a single semicolon) without any effect on your program.

It has the meaning of an a statement followed by an empty statement.
In C each statememnt ends with ;. So a statement with a ; followed by one, is a statement followed by an empty statement.

A "double semicolon" does not have any special meaning in c. The second semicolon simply terminates an empty statement. So you can simply remove it.

Most likely a typo. Adds a null statement to your program.

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I was trying to declare a function which takes a string as an input and give an int as an output [closed]

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my code is not identifying type string. I am using c to program
In your calls to count_letter(), count_words() and count_scenteces() you include a unnecessary string type in the function argument.
You should only include string when declaring a variable of some sort for example:
string x = "hi";
So to fix this problem replace each instance of string text in your function calls with text.
This will reference the text variable instead of inappropriately introducing a new variable of type string into the program.
Parameter types are only put in function declarations, not function calls.
int letters = count_letters(string text);
should be
int letters = count_letters(text);

Why the output of printf("%d","") is 173 in c language? [closed]

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#include<stdio.h>
#include<conio.h>
void main()
{
printf("%d","");//printing output
getch();
}
The output is 173 .I am not getting why the output is 173.
First, you're trying to print a string as decimal integer, which means the decimal you try to print is going to be the pointer to the string (actually a pointer to the array of characters) and not the string itself. To use an individual character use single-quotes, not double-quotes.
To accomplish what you're actually trying to do, do this:
printf("%d", ' ');
Note there is an actual space between the two single-quotes.
The result will be 32, which is the decimal value for the ASCII Space character.

expected expression before -return value- [closed]

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Given the following function:
image_ret* minify_1(image_src img_src, CLIENT* cl) {
image_ret* img_ret;
magickminify_init();
magickminify(img_src.image_src_val, img_src.image_src_len, (ssize_t*)&img_ret->image_ret_len);
return image_ret;
}
The compiler is telling me "expected expression before ‘image_ret’" with regard to the last line. I'm sure I'm missing some fundamental aspect of syntax here, but I don't know what. Lil' help?
You need to return a value, not a type. image_ret is a type, img_ret is a poitner to a value of that type and probably what you want to return, except I see nowhere in your code where you allocate any storage to it, or initialising any of the fields except image_ret_len

How can I detect my coding mistake when I receive this message [closed]

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link to the code: http://gyazo.com/f0f4004eb606607ecaa021b5e22e6e06
I am getting the following error when i am running th code.
"error: expected identifieror '(' "
I use gedit to write this code.
I would appreciate some support guys ;)
Thanks in advance!,
Vicente
There's shouldn't be a semicolon in the int main(void); declaration.
Try replacing line 4 with: int main(void) instead.
Also, please read up on C function declaration syntax
https://msdn.microsoft.com/en-us/library/c4d5ssht.aspx.
int main(void);
^you should not do this.
And you forgot to put ; after this statement -
int height=n
And also n is not declared in your program.

How to terminate while loop if character matched '.' in c [closed]

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char b;
while((b=getchar()) != '.' );
{
printf("%c",b);
}
If I had the following input in stdin abcd.
it should print a then b then c then d then detect the . and terminate although its simply printing a . instead of abcd
Remove the semicolon at the end of while condition
The semicolon at the end of the while loop makes it execute only once . It does not loop at all . Your program is simple a linear program with no looping .
Otherwise your program logic is fine .

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