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How can we replace the value of const variable using scanf statement?
Wiki User
∙ 13y ago
You cannot change the value of a const variable. That's what const means - it is constant.
If you are considering "trickery" using pointers and the scanf function, understand that this is not supported, is highly non portable, and may fail, depending on whether or not the implementation places its const data in read only memory. Besides, most modern compilers will not allow you to place a const variable as a non-const argument to a function.
Use the language within its defined boundaries.
Wiki User
∙ 13y ago
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What is the full form of printf or scanf?
int printf (const char *fmt, ...)orint scanf (const char *fmt, ...)
What happens by the 'scanf' statement in C programming language?
By using the scanf statement in c,we can take the values at run time . for ex a is a variable of type int so, we can assign the value of a using scanf at run time as follows. scanf("%d",&a); here %d is the conversion specifier for integer type.
What is the syntax of scanf?
scanf: scanf is a built-in function to read input from the user.
Why not use the ampression symbol in scanf statement in function?
scanf is a function, not a statement. Example: int a; char name[12]; scanf ("%d %s", &a, &name[0]);
What Program structure for a C program?
1.Declaration of header files. 2.void main()/main() 3.delcaration of the variable. 4.Printf&scanf statement. 5.execution. 6.result
What happens by the scanf statement in c?
It's a function, not a statement. Use your help system to find out its purpose.
What is the purpose of the scanf function?
The scanf() function reads formatted input from stdin according to a format string and assigns the data to specified variable(s). The input data is a character sequence which is converted to the appropriate type, as per the format string, before being assigned to the corresponding variable. Examples: unsigned x; scanf("%u", &x); In the above example, we read an unsigned integer (%u) from the stdin character stream and assign the resultant value to the variable x.
When do we use an address operator in c'?
In C we use & operator while giving address of some variable to some pointer variable. & operator is also used in scanf().
What is the prototype of scanf in c programmin?
int scanf(char* format, ...); the format accepts the format specifier string, the elipsis operator accepts the variable list scanf("var type var type ...", &var, &var, ...); example: int num1, num2, num3; scanf("%d %d %d",&num1,&num2,&num3); Is that what you were looking for? Maybe this can help also...
What is the use of scanf function in c?
Hiii,scanf() is an predefined function of C to getting runtime input from the user.Syntax:-scanf("sequence specifier",&variable_name);example:-#includevoid main(){int number;printf("\n Enter any number"); //message to userscanf("%d",&number); //getting input from userprintf("\n entered number is %d",number); // displaygetch();}
Why you use scanf?
scanf is a function that reads data with specified format from a given string stream source, originated from C programming language, and is present in many other programming languages.The scanf function prototype is:int scanf(const char *format, ...);The function returns the total number of items successfully matched, which can be less than the number requested. If the input stream is exhausted or reading from it otherwise fails before any items are matched, EOF is returned.So far as is traceable, "scanf" stands for "scan format", because it scans the input for valid tokens and parses them according to a specified format.UsageThe scanf function is found in C, in which it reads input for numbers and other datatypes from standard input (often a command line interface or similar kind of a text user interface).The following shows code in C that reads a variable number of unformatted decimal integers from the console and prints out each of them on a separate line:#include intmain(void){int n;while (scanf("%d", &n) 1)puts(word);return 0;}No matter what the datatype the programmer wants the program to read, the arguments (such as &nabove) must be pointers pointing to memory. Otherwise, the function will not perform correctly because it will be attempting to overwrite the wrong sections of memory, rather than pointing to the memory location of the variable you are attempting to get input for.As scanf is designated to read only from standard input, many programming languages with interfaces, such as PHP, have derivatives such as sscanfand fscanf but not scanf itself.
How do you read a character from the keyboard and will store it in the variable?
Use the scanf() function from the C standard library.
