I'm trying to make a pointer array which in my case can point to three elements.I want to take the elements at run time and want to access each element but I couldn't. Here is the source code.
#include<stdio.h>
main(){
int i;
float *coefficient[3];
for(i=0;i<=2;i++){
scanf("%f",coefficient[i]);
}
for(i=0;i<=2;i++){
printf("%f\n",*coefficient[i]);
}
}
Change
float *coefficient[3];
to
float coefficient[3];
As you need an array of float not float pointers
Then,
change
scanf("%f",coefficient[i]);
to
scanf("%f",&coefficient[i]);
As you need a pointer to the float to read in. Please also check the return value from scanf
As you have an array of floats
printf("%f\n",coefficient[i]);
is what you require
Your pointers are uninitialized, so dereferencing them causes undefined behavior. You need to allocate memory for them, e.g. using malloc().
for(i=0;i<=2;i++){
coefficient[i] = malloc(sizeof *coefficient[i])
scanf("%f",coefficient[i]);
}
And after you print them, you should use free() to release the memory.
For starters according to the C Standard the function main without parameters shall be declared like
int main( void )
You declared an array of three pointers but the elements of the array are not initialized and have indeterminate values.
As a result calling the function scanf in this loop
for(i=0;i<=2;i++){
scanf("%f",coefficient[i]);
}
invokes undefined behavior.
The pointers which are elements of the array shall point to a valid memory where data may be stored using the pointers.
For example you can for each pointer dynamically allocate memory that will be pointed to by the corresponding pointer
Here is a demonstrative program.
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
enum { N = 3 };
float *coefficient[N];
for ( size_t i = 0; i < N; i++ )
{
coefficient[i] = malloc( sizeof( float ) );
}
for ( size_t i = 0; i < N; i++ )
{
scanf( "%f", coefficient[i] );
}
for ( size_t i = 0; i < N; i++ )
{
printf( "%f\n", *coefficient[i] );
}
for ( size_t i = 0; i < N; i++ )
{
free( coefficient[i] );
}
return 0;
}
If to enter
1.1
2.2
3.3
then the output will look like
1.100000
2.200000
3.300000
You can format the output as you like.
In general you should check that a call of malloc and/or of scanf was successful.
int main() {...} than ever before, and here OP code doesn't seem to acknowledge main() returning a value at all. Hmmm. Maybe it's a retro computing thing. IAC, it might be helpful to point out that more robust code checks for failure of malloc() and checks the value returned by scanf().
coefficientis already essentially a pointer. Try declaringfloat coefficient[3];instead offloat *coefficient[3]*in theprintf, and change thescanftoscanf("%f", &coefficient[i]);to match. I suspect they declared a pointer array solely because they didn't understand how you're supposed to pass arguments toscanf(using&to get the address of an existing stack variable typically).