英文:
How to resolve the segmentation fault issue while working with pointers in c?
问题
#include<stdio.h>
void main(){
float p1, p2;
float *x, *y;
printf("输入坐标 (x, y)：");
scanf("%f%f", &p1, &p2);
x = &p1;
y = &p2;
printf("坐标(%f,%f)位于象限 ", *x, *y);
if (*x == 0 && *y == 0) {
printf("它是原点。");
return;
}
if (*x > 0 && *y > 0) {
printf("1");
}
else if (*x < 0 && *y > 0) {
printf("2");
}
else if (*x < 0 && *y < 0) {
printf("3");
}
else printf("4");
return;
}
英文:
#include<stdio.h>
void main(){
float p1,p2;
float *x,*y;
printf("Enter the co-ordinates (x,y) : ");
scanf("%f%f",p1,p2);
x = &p1;
y = &p2;
printf("The co-ordinates(%f,%f) lies in quadrant ",*(x),*(y));
if(*x == 0 && *y ==0) {
printf("It is the origin.");
return;
}
if(*x > 0 && *y > 0){
printf("1");
}
else if(*x < 0 && *y > 0){
printf("2");
}
else if(*x < 0 && *y < 0){
printf("3");
}
else printf("4");
return ;
}
This code is throwing an error "Segmentation Fault".
can anyone help me out ?
I have tried to assign values to normal variables then pointers are assigned to that variable's address.But yet it is throwing the same error.
答案1
得分: 3
以下是要翻译的内容：
The line
scanf("%f%f", p1, p2);
is wrong. %f in scanf() expects pointers of float variable, so it should be
scanf("%f%f", &p1, &p2);
Checking for input failure is better.
if (scanf("%f%f", &p1, &p2) != 2){
puts("failed to read values");
return;
}
英文:
The line
scanf("%f%f",p1,p2);
is wrong. %f in scanf() expects pointers of float variable, so it should be
scanf("%f%f",&p1,&p2);
Checking for input failure is better.
if (scanf("%f%f",&p1,&p2) != 2){
puts("failed to read values");
return;
}
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