I used Eclipse to compile c code, but all of a sudden all my codes got trouble, which were all correct previously.
for example if I want use scanf input a argument, before the scanf a printf statement I will use for guiding the user. like printf("type the size\n"); but after compiling in Console I need type the size first, then the printf("type the size\n") command just pop up, which should be the other way round.
#include <stdio.h>
#include <stdlib.h>
void try(int a);
int main(void)
{
int a;
printf("type the size\n");
try(a);
return 0;
}
void try(int a)
{
scanf("%d", &a);
printf("%d\n", a);
}
the result:
2
type the size
size is chosen 2
I need type a number first, here like I need type 2 first and then the "type the size" just pop up.
here is what I want :
type the size
2
size is chosen 2
Its a bug in eclipse and this has been reported by most of the people using eclipse and MinGW.
To overcome this problem, you could use fflush(stdout) after every call to printf or use the following in the start of main :
setvbuf(stdout, NULL, _IONBF, 0);
setvbuf(stderr, NULL, _IONBF, 0);
This will cause stdout and stderr to flush immediately whenever it is written to.
try this one, for example,
#include <stdio.h>
#include <stdlib.h>
void try(int *a);
int main() {
int a;
printf("type the size\n");
fflush(stdout);
try(&a);
return 0;
}
void try(int *a){
scanf("%d", a);
printf("%d\n", *a);
fflush(stdout);
return;
}
Also, you need to get the value set in atry back in main, so, you need to pass a a pointer as shown
Related
void pedir_diag(int* diag){
scanf("%i", &diag);
}
int main() {
int number;
pedir_diag(&number);
return 0;
}
When I compile introduce an integer and expect that number int variable in the main have the value introduced but is not set
I have this, in my code but i am not able to set diag variable with scanf() function.
diag is already a pointer so no need for the address-of operator (&) in the call to scanf():
void pedir_diag(int *diag)
{
scanf("%d", diag);
}
But to do it that way is kinda stupid. You have no way to check for erroneous input.
Better:
#include <stdbool.h>
bool pedir_diag(int *diag)
{
return scanf("%d", diag) == 1; // *)
}
*) scanf() returns the number of successful conversions, so if the number of successful conversions is 1 we return true.
Usage:
#include <stdio.h>
int main(void)
{
int foo;
if (!pedir_diag(&foo)) {
fputs("Input error :(\n\n", stderr);
}
else {
// have fun with foo
}
}
The best way to use scanf through your entire coding life is by remembering the following rule
scanf("%[direction_of_what_to_scan][special format character]",
Pointer_on_some_memory_location);
What scanf does is that it stores what the input was (with some length restrictions) to a memory address. So, either:
int n1;
scanf("%d",&n); //which means the address of variable n1 in memory
or
int *n2 = &n1;
scanf("%d",n) // &doesn't need to be used cause n2 is now already pointer to an integer
Both will are different implementations of the same thing, they differ to the part of using pointers,for which C is well-known,and even applicable these days.
I just started with C and am tasked with using a header to house a prototype for a function. The problem is that nothing happens when I'm expecting a prompt for input. I didn't get an error and would like to know where to look at first to solve my problem. This is what I have so far.
LAB2.c
#include <stdio.h>
#include "LAB2HEADER.h"
int main(){
double *p;
double array [10];
p = array;
const int size = 10;
void input(p,size);
return 0;
}
LAB2HEADER.h
#ifndef LAB2HEADER_H_
#define LAB2HEADER_H_
void input (double *array,const int size);
#endif
LAB2HEADER.c
#include <stdio.h>
#include "LAB2HEADER.h"
void input (double *array,const int size){
for (int i = 0; i < size ; i++)
{
printf("Input a value");
scanf("%lf", &array[i]);
}
}
A lot of the notes I look at seem to only either use Int as a parameter or have a function with no needed parameters, could my mistake be in my array pointer is it a problem with the way I made my function?
void input(p,size);
This line makes no sense. If this is supposed to be a function call, you need to remove void.
Also, since your print statement does not end with a newline, nor do you flush stdout before reading in the value, your prompt might still be in the output buffer and not be output until you hit the newline AFTER entering the value.
I am very new to C programming and having trouble compiling what should be a very simple function. The function, called printSummary, simply takes 3 integers as arguments, then prints some text along with those integers. For example, if hits=1, misses=2, and evictions=3, then printSummary(hits,misses,evictions) should print the following:
hits:1 misses:2 evictions:3
Here is the code I'm using. Thanks in advance for any advice.
#include<stdio.h>
void printSummary(int hits, int misses, int evictions)
{
printf('hits: %d\n');
printf('misses: %d\n');
printf('evictions: %d\n');
}
int main()
{
int hit_count = 1;
int miss_count = 2;
int eviction_count = 3;
printSummary(hit_count, miss_count, eviction_count);
return 0;
}
Compiling this code gives me several warnings, but no errors. When I run the code, I get a segmentation fault. Like I said, I am fairly new to C so there is most likely a simply solution that I am just missing. Thanks in advance for any advice.
Make the below changes .
printf("hits: %d\n",hits);
printf("misses: %d\n",misses);
printf("evictions: %d\n",evictions);
printf has a
int printf(const char *format, ...)
prototype. So in the first argument you can pass format specifiers and in the next provide the actual variables/values to be printed out
errors are:
void printSummary(int hits, int misses, int evictions)
{
/* Name: printf
Prototype: int printf (const char *template, ...)
Description:
The printf function prints the optional arguments under the
control of the template string template to the stream stdout.
It returns the number of characters printed,or a negative value if
there was an output error.*/
printf("hits: %d\n", hits); // don't use ' it is used only for char variable for example: char a = 'c';
printf("misses: %d\n", misses);
printf("evictions: %d\n", evictions);
}
Your printf function is not being called correctly. You have to include the integers needed to print:
printf("hits: %d\n", hits);
printf("misses: %d\n", misses);
printf("evictions: %d\n", evictions);
Read more about the printf function here.
I am just starting to learn programming for a unit I am doing in my engineering course and I have come across pointers. I just wanted some reassurance that I actually understand the concept correctly in terms of using a pointer as an argument in a function. If I understand it correctly, you pass a pointer to an address of a variable you would like to be altered by a separate function called, even though it is a local variable within the scope of the calling function. Does that make sense? I have an example from my text book which I re-wrote. The only thing is they gave it in two incomplete parts and I put it together, filled in the blanks and added the final printf statement in the main function. I'll paste it here:
#include <stdio.h>
#include <stdlib.h>
#define READ_OK 0
#define READ_ERROR 1
int read_num(int lo, int hi, int *num);
int main(int argc, char *argv[])
{
int lo = 0, hi = 0, *num, val;
printf("Please enter a lower bound and an upper bound for your range,respectively\nLower: ");
scanf("%d", &lo);
printf("Upper: ");
scanf("%d", &hi);
num = &val;
if(read_num(lo,hi, &val) != READ_OK)
{
printf("Read error, program abort\n");
exit(EXIT_FAILURE);
}
else
{
printf("You entered %d, press any key to continue: \n", val);
getchar();
}
return 0;
}
int read_num(int lo, int hi, int *num)
{
int next;
printf("Enter a number between %d and %d: ", lo, hi);
while(scanf("%d", &next)==1)
{
if (lo<=next && next<=hi)
{
*num = next;
return READ_OK;
}
printf("%d is not between %d and %d\nTry again: ", next, lo, hi);
}
return READ_ERROR;
}
So is my understanding correct? "val" gets modified in read_num() by passing it's address in the form of pointer "*num", in which the the value for "next" is then written?
PS: is this syntax correct?
PPS: What would this process specifically be called?
Thanks a bunch for any help :)
The *num is not necessary inside the main() function. As you are passing the address of the val inside the read_num() , so any changes from the read_num() will also affect the value inside main() as you are working with the address.
In your program you have basically use two different pointers- one is inside main which is num, and another inside read_num() which is also num, for more understanding see the scope of a variable in c. As the val is inside main so you don't need to use pointer here, because you have the access of changing the value from the main as it is local to it. You will need pointer when you will be changing the value of val outside from the main, or from outside of the scope of the variable.
i have written this code for a question on codechef (A4)....when i give the input:
2
4 2
This program stops unexpectedly without taking further input ....can some please point out the mistake in the code?
#include <stdio.h>
#include<math.h>
void number(long int a,int b)
{
int c;
c=b;
int first[c],last[c],e=1,i;
long int d;
d=pow(a,a);
for(i=(c-1);i>=0;i--)
{
last[i]=fmod(d,pow(10,e));
e++;
}
e=1;
while(d>pow(10,(b-1)))
d/=10;
for(i=(c-1);i>=0;i--)
{
first[i]=fmod(d,pow(10,e));
e++;
}
for(i=0;i<c;i++)
printf("%d",first[i]);
printf(" ");
for(i=0;i<c;i++)
printf("%d",last[i]);
printf("\n");
}
int main()
{ int T;
scanf("%d",&T);
while(T--)
{ long int a;
int b;
scanf("%ld %d",a,b);
number(a,b);
}
return 0;
}
scanf("%ld %d",&a,&b);
Using uninitialized variables lead to UB. You should use &a and &b to scan variables
In your code you have
scanf("&ld %d",a,b);
It means you're trying to input integers to the memory locations of values of a and b. For and example let value of a = 1234566466 (long int), and b = 1234 (int). Accordingly 1234 is a memory location which is at the start of the RAM. Tn that area System files are loaded. So you are going to change system behaviour. That is not allowed.
Further when the complier allocate some memory space to your program, you can access only the memory which is inside your memory segment directly. But above statement trying to access another segment.
That's why you get segmentatin fault.
You are passing an integer to a function that expects a pointer, for scanf the "%d" and "%ld" specifiers expect int * and long int * respectively, and you pass int and long int, so when trying to access the integers as if they were memory addresses the segmentation fault occurs.
The correct way to call scanf would be as Gopi said
scanf("%ld %d", &a, &b);
there you pass a and b addresses instead of their values.