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# Can anyone explain this code

// #include<stdio.h> int main() { int var1= 4/9; printf("%d\n",var1); float var2=4/9; printf("%.2f\n",var2); float var3=4.0/9.0; printf("%.2f\n",var3); return 0; }

28th Jun 2022, 3:33 PM
JAGAN GANIREDDY
2 Respostas
+ 3
JAGAN GANIREDDY Dividing float(or double) by float(or double or int ) will result float like 4.0/9.0 =0.44444// float by float 4.0/9 =0.44444 //float by int 4/9.0=0.44444 // int by float but if we divide int by int the compiler will ignore the decimal part like this 4/9 = 0 not 0.44444 5/2 = 2 not 2.5 printf("d%",var1);// 0 printf("0.2f%",var2);// 0.0 printf("0.2f%",var3);// 0.44 as we use 0.2% this will print 2 values after the decimal point
28th Jun 2022, 3:44 PM
Aly Alsayed
+ 1
Mathematically, 4/9 ā 0.4444444 In C Language, 4/9 = 0 4/9.0 ā 0.444444 4.0/9 ā 0.444444 4.0/9.0 ā 0.444444 (float) 4/9 ā 0.444444 (int) 4.0/9 = 0 (int) 4/9.0 = 0 So, in 'float var2 = 4/9;' the value will be 0.0 as 4/9 is 0 but in 'float var3 = 4.0/9.0' the value will be 0.44
28th Jun 2022, 3:58 PM
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