Programmer --- C language details 20 (Conversion between symbols and symbols, and calculation after two numbers overflow)

Programmer --- C language details 20 (Conversion between symbols and symbols, and calculation after two numbers overflow)
Main Content: unsigned and converted between symbols, numeric calculation after two numbers Overflow

# Include
 
  
/* This function has a potential vulnerability */float sum_elements (float a [], unsigned length) {int I; float result = 0; for (I = 0; I <= length-1; I ++) {result + = a [I]; printf ("a [% d] = % f \ n", I, a [I]);} return result;} int main () {int I = 200*300*400*500; // int indicates about 2 billion, unsigned, about 4 billion float j = (3.14 + 1e20)-1e20; // The output result is 0 due to limited precision, and the output of this statement is 3.14: float j = 3.14 + (1e20-1e20); // e indicates that 10 is the low index long int a = 1; long int B = 10; printf ("I = % d \ n", I); printf ("j = % f \ n", j); printf ("a = % ld \ n ", a); printf ("B = % ld \ n", B); printf ("\ n");/* test the completion Code */unsigned int u = 4294967295u; int tu = (int) u; printf ("u = % u, tu = % d \ n", u, tu ); // The maximum value of the unsigned int is the same as that of-1 (that is, the same bit of the unsigned Umax and-1) printf ("\ n "); // when a signed number is mapped to its unsigned number, the negative number is converted to a large positive number, and the non-negative number remains unchanged/* test conversion */short int v =-12345; unsigned short uv = (unsigned short) v; // force conversion changes the value without changing the bit representation (that is,-12345 is the same as the unsigned 53191 bit) printf ("v = % d, uv = % u \ n", v, uv); printf ("\ n"); float c [3]; sum_elements (c, 3); // an error occurs when 0 is passed. There is a problem of unsigned and Signed Conversion./* conclusion: one number of expressions is signed, the other is unsigned, And the C language is converted to unsigned by default, when comparing-1 <0U, there will be a problem * // * for the overflow calculation formula, see */short a1 =-65536; short b1 =-1; printf ("\ n "); printf ("% d \ n", a1 + b1); return 0 ;}
 

Overflow formula:

Output: