As described in the heading, the terminal delivers the error: zsh: segmentation fault. Here is the - pretty basic - code:
#include<stdio.h>
int main(){
int age = 0;
printf("Input your age!");
scanf("%d", &age);
printf(age);
}
thanks for the help solving this problem :)
printf(age);
should be printf("%d", age);
You need to pass in a string literal (eg: "hello world") with a format specifier (exactly the same way you did it in scanf) and then pass the age (not its address as you did with scanf, so without the &) as a second argument.
for example
printf("I am %d years old\n", age);
For the above example, I printed a message that writes "I am [age] years old" and then continues down to a new line (that's the '\n'). The '%d' is a format specifier; what it does is it specifies that in its position, the function should print a value of a specific type (%d specifies an integer). You can have multiple of these specifiers and it prints the arguments you give it in linear order. I would highly recommend you look at this[1], it should explain it better than I can.
Related
Just like printf(), I was trying to use optional specifiers in scanf() format string. I tried to use the width and precision specifier. Now in printf() it simply reserves the columns and print according to these specifiers but what happens with scanf()? Like what is meaning of %3d and %3.3f here? Is this even relevant in case of scanf()? I have a little idea that width in this case represents the number of characters that are to be read for some particular format but not sure. Below code explains this further:
#include<stdio.h>
int main()
{
int a;
float b;
printf("Enter Numbers:\n");
scanf("%3d %3.3f",&a,&b);
printf("Entered Numbers are\n");
printf("%d %f",a,b);
return 0;
}
Since you specified in the comments that what you really want to know is 'what if i forcefully try to do it' ... Here are the results (with Clang)
warning: invalid conversion specifier '.'
and
warning: data argument not used by format string
The program compiles , however, since these are just warnings.
Upon executing the binary, and entering the variables asked for:
The "%d" for a gets stored properly.
Regardless of what value is entered, the " %3.3f " for b always stores 0.000000
In short, the it does what almost any other code that compiles with warnings does - not behave as intended. This is neither undefined, nor unspecified behaviour, but it is wrong.
Suggestion : Refrain from asking questions that are of the nature ' what happens if I try to compile this '. Just try and see for yourself !
I am writing a simple C program to accept an input number from user and display it just because earlier i was writing some C program and this stupid error is bugging me it wasn't there before until yesterday
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int a;
printf("Enter a number");
scanf("%d",&a);
printf("Display number%d",&a);
}
Every time i run a program that accept an input it displays a seemingly random value and not the one I entered not just in this program in any other too, here
is the O/p:
Enter a number: 12
Display number : 2752300
Process returned 7(0X7) execution time : 1.880s
Press any key to continue
I don't know whether is the Compiler error or some memory error that is causing this problem but for the record i have tried using different IDE like DEV C/C++, Turbo C/C++ and Code Blocks but the error remains to be same in all of it except in Turbo C/C++ it display a signed number of i enter for eg : if input is 12 it displays: -12.
The problem is in your print statement. You are printing the address of a, not the value. Use print("%d",a);
The %d format specifier for printf expects an int, not an int *:
printf("%d",a);
It's not attempting to write to a, so it doesn't need its address.
If you compile with the warning level turned up (-Wall -Wextra for gcc), it will tell you this:
format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
scanf needs a pointer to the variable if it is going to change the variable itself, so we use the address-of operator to get the pointer.
scanf() received "%d", as a parameter and then knows it needs to read an integer and store it in the next variable in the argument list.
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
int a = 0;
printf("Enter a number: ");
scanf ("%d" ,&a);
printf ("%d \n ",a);
return 0;
}
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 7 years ago.
Improve this question
I have started learning C and I want to make an addition program.
I have written the following code:
#include <stdio.h>
#include <stdlib.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
int main() {
int a,b,c;
scanf("Give a,b",&a,&b);
c=a + b;
printf("A+b=",c);
printf("\n\n\n");
return 0;
}
It is supposed to take 2 numbers from the users and show their sum.
However the only output I get after the user writes 2 numbers is A+b=
Any ideas?
You failed to include a format specifier for your integer variable c. Without this printf doesn't know what arguments to expect following the format string, what their types are, or how they are to be printed. Change:
printf("A+b=",c);
to:
printf("A+b=%d",c);
Note that a good compiler with warnings enabled (e.g. gcc -Wall ...) would have pointed out this simple mistake for you at compile-time.
Also your scanf usage is wrong - change:
scanf("Give a,b",&a,&b);
to:
printf("Give a,b");
scanf("%d %d",&a,&b);
You're missing the %d conversion specifier in the printf() call:
printf("a+b=%d\n", c);
Without that, printf() doesn't know it's getting a second argument and won't do anything with it.
Of course, that c could just be a + b, there's no need to store the sum in a separate variable just to print it.
Also, you need specifiers in scanf(), it should be:
scanf("%d %d", &a, &b);
The first argument to scanf() is not a prompt that's printed, it's describing the expected input.
Last, you should check the return value of scanf() to make sure it's 2 before relying on a and b having valid values. I/O can fail, you need to make sure your program does the right thing if that happens.
scanf is a function used to take in input from the user. You are expecting it to print something, ain't you? You need printf instead:
printf("Give a,b\n"); //\n at the end is good
Now, use scanf to scan in the input:
scanf("%d %d",&a,&b);
You are also missing the %d format specifier in the last printf as other answers have mentioned.
scanf() is an input C function (scan with formatting) that uses format specifiers (%datatype ex: %d or %i = int, which is what you want, or %c = char) to store a data entry in given format, provided by the stdin (keyboard buffer) up to and including a null (\0) termination char(in the case of a string);
The Ampersand (&) in C is used to designate/return the memory address (where variable is stored), as opposed to the value of the variable.
scanf("Give a,b",&a,&b);
c=a + b;
printf("A+b=",c);
The issue with above first line of code is that you forgot to include the format specifiers (%d or %i would both work in this case, as I assume you are adding integers and not floats/etc).
Thus the solution to your first problem is an easy one:
scanf("%d %d",&a,&b);
Also, it seems that you are attempting to combine an input and output in one line.
Printf, like scanf, can use format specifiers, or it can contain only a string of characters, and prints to the stdout (console).
What you want to do is include a prompt asking the user for input before storing said input. You add the \n "newline" character at the end to ensure to provide spacing between other outputs and inputs, among many other reasons when you get into char arrays[], aka strings. But for now, this should do the trick:
printf("Give a,b\n");
scanf("%d %d",&a,&b);
Hope this helps both with the current assignment and with general/C programming concepts!
Here you can see my source code:
#include <stdio.h>
int main()
{
char yourname;
char yoursex;
int yourage = 0;
printf("Hey, what's your name?\n");
printf("My name is: ");
scanf("%s", &yourname);
printf("Oh, hello %s! \n\n", &yourname);
printf("Are you a boy or a girl?: ");
scanf("%s", &yoursex);
printf("Nice to know you are a %s! \n\n", &yoursex);
printf("How old are you %s? I am ", &yourname);
scanf("%d", &yourage);
printf("I see you are %d, you have many years then!", &yourage);
return 0;
}
I was trying things that I didn't knew, and strangely it is not working for me. What's the problem? Also, why it needs to be %s and not %c? If I use %c instead it does not work!
Where it says:
How old are you %s? instead of putting my name, it says ''oy''
and instead of showing my age in the last line, it shows a big number.
These are the very basics of C Programming, and I strongly advise you to get a decent book - The C Programming Language by Dennis Ritchie would be a good start.
There are numerous errors in your code.
A char can contain only one character, like 'A', or 'a' or something like that. When you're scanning a name, it is going to be a group of characters, like 'E', 'd', 'd', 'y'. To store multiple characters, you need to use a character array. Also, the format specifier used to scan/print characters is %c, %s is for when you need to scan a group of characters, also called a string into an array.
When you use printf, you do not supply a pointer to the variable you are trying to print (&x is a pointer to variable x). The pointer is a 32/64-bit integer, which is likely why you see a random integer when trying to print. printf("%c\n", charVar) is sufficient.
scanf does not need an & while using %s as the format specifier, assuming you have passed a character array as the argument. The reason is, scanf needs to know where to store the data you are reading from the input - and that is given by a pointer to the memory location. When you need to scan an integer, you need to pass an &x - which means, pointer to memory location of x. But when you pass a character array, it is already in the form of a memory address, and doesn't need to be preceded by an ampersand.
I once again recommend you look up some decent tutorials online, or get a book (the one I mentioned above is a classic). Type the examples as given in the material. Experiment. Have fun. :)
%s is for reading a string -- multiple characters delimited by whitespace. %c is for reading a single char.
You declare your yourname and yoursex vars as characters, and then try to read strings into them. The string read will overwrite random other things in the stack frame and misbehave or crash.
You want to declare yourname and yoursex as character arrays, so they can hold strings:
char yourname[32];
char yoursex[32];
then, when reading into them, you want to include a length limit so they don't overflow:
scanf("%31s", yourname);
This is a single character:
char yourname;
But %s indicates that the variable is a string (i.e., an array of characters terminated by a NUL). That's why you need %c. If you really did mean to use a string, then define the variable like
char yourname[32]; /* just pick a big enough size */
Also, you are correct to use the address of the variable with scanf(), but printf() needs the value. So instead of
printf("I see you are %d, you have many years then!", &yourage);
use
printf("I see you are %d, you have many years then!", yourage);
The "big number" is the memory address.
Make sure you read the comments in code!
#include <stdio.h>
int main()
{
char yourname[10];
char yoursex[5]; // boy or girl + null terminator
int yourage = 0;
printf("Hey, what's your name?\n");
printf("My name is: ");
scanf("%s", &(*yourname)); // & and * cancel each other out,
// thus take a look at the next scanf()
printf("Oh, hello %s! \n\n", yourname); // yourname is now an array
printf("Are you a boy or a girl?: ");
scanf("%s", yoursex);
printf("Nice to know you are a %s! \n\n", yoursex);
printf("How old are you %s? I am ", yourname);
scanf("%d", &yourage); // ok
printf("I see you are %d, you have many years then!", yourage); // here you don't
// need the address of the variable!
return 0;
}
The expression char yourname; only holds space for a single character, so quite likely you end up corrupting the memory space when scanning for yourname. You should allocate a bigger buffer and make sure that you don't overrun its length by setting a maximum number of characters to be read with the scanf function; as described in some of the other answers.
The fact that the following printf print correctly the name doesn't mean that the memory doesn't get corrupted; as C/C++ don't really check the boundary of any strings or arrays used at runtime.
As suggested by others, starting by reading a good book about C and/or C++ wouldn't a bad idea.
I need help on this exercise from C Primer Plus.
Write a program that requests your first name and does the following with it:
Prints it in a field three characters wider than the name
#include<stdio.h>
#include<string.h>
int main()
{
char a[40];
int p,v=0;
printf("Enter your first name: \n");
scanf("%s",a);
p= strlen(a);
v==p+3;
printf("%s",a);
}
I cant figure out how what to use as a modifier for the width
what should I add in between % and s?
the goal of this excercise is to read and grok the manual page for printf(). The reading part could only be done by you and there is no shortcut. The format specifier is the most complex chapter in C-Programing (other would say 'pointer'), and it is very wise to know where look things up (man-page) in need of remembering.
When you are done reading, you should have a little (or big) understanding of the format-specifier %s with all its possibilities.
EDITH: when you are done reading, and there is still a question whether to use "%*.*s" or "%s" or "%-.*s" etc., please come back with an updated question.
Here is one way to do it:
#include<stdio.h>
#include<string.h>
int main()
{
char a[40];
printf("Enter your first name: \n");
scanf("%s",a);
printf("[%-*s]", (int)(3 + strlen(a)), a);
return(0);
}
The printf() function can do it all.
From the question code, it is clear that "%s" as a format string is understood.
A format string of "%*s" allows the caller to place the (space-padded) width to be specified. For example:
printf("%*s", 10, "Hello");
The above will print "Hello" (as expected), in a 10-character frame. Hence, the command above actually prints: " Hello".
To put the spaces on the other side, tell printf() to left-justify the string using:
printf("%-*s", 10, "Hello");
This results in printing: "Hello "