What does C or GCC do when implicit function declaration exists

#include <stdio.h>int main(void){  float a, b;  a = 0.0f, b = 1.0f;  printf("%d\n", plus(a, b));  return 0;}int plus(double a, double b){  printf("%lf %lf\n", a, b);  return a + b;}

gcc .\p1.c --std=c89 -o main

the output is:

0.000000 1.0000001

But when i changed the type from float to int

#include <stdio.h>int main(void){  int a, b;  a = 0, b = 1;  printf("%d\n", plus(a, b));  return 0;}int plus(double a, double b){  printf("%lf %lf\n", a, b);  return a + b;}

gcc .\p1.c --std=c89 -o main

the output is:

32019693379112340.000000 622696491526558860000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000.000000-2147483648

In the first stage, the default argument promotion changes float to double, and everything is right.

In the second, int -> unsigned int, why does it also call plus function i defined, and why the value of argument in the plus function is out of expectation

I tried to use following to simulate the process:

#include <stdio.h>int main(void){  unsigned int a1, b1;  a1 = 0u, b1 = 1u;  double da = a1;  double db = b1;  printf("%d\n", plus(da, db));  return 0;}int plus(double a, double b){  printf("%lf %lf\n", a, b);  return a + b;}

gcc .\p1.c --std=c89 -o main

the output is:

0.000000 1.0000001

What confuses me even more is this:

#include <stdio.h>int main(void){  unsigned int a1, b1;  a1 = 0u, b1 = 1u;  double da = a1;  double db = b1;  printf("%d\n", plus(da, db));  int a, b;  a = 0, b = 1;  printf("%d\n", plus(a, b));  double x = 3.0;  printf("%d\n", add5(x));  return 0;}int plus(double a, double b){  printf("%lf %lf\n", a, b);  return a + b;

gcc .\p1.c --std=c89 -o main

the output is:

0.000000 1.00000010.000000 0.0000000