Hi ,

I have completed a programming. Just wondering if anyone have any idea on how to reduce the source code to a more simple one(display part, red color). For this program , i check for whether there is any hexadecimal 10~15. if yes, i display accordingly.

That mean i am now using 16 "if" function. Not a very tidy programming.

Anyone with a much better idea ???

Code:#include<stdio.h> main() { int i,j,ch,decimal,count,search; int temp; int hex[10]; int number[10]; i=0; count=0; decimal=0; printf("\nPlease enter your input in decimal form: "); ch = getchar(); while(ch != '\n') { if('0' <= ch && ch <= '9') { decimal = decimal * 10; decimal = decimal + (ch - '0'); } ch = getchar(); } printf("\nInput? %d\n",decimal); while(decimal != 0) { hex[i] = decimal%16; decimal= decimal/16; i++; count++; } i=i-1; for(j=0; j < count; j++,i--) { number[j]=hex[i]; } printf("\nThe hexadecimal is : "); for(j=0; j < count; j++) { if(number[j]==0) {putchar('0');} else if(number[j]==1) {putchar('1');} else if(number[j]==2) {putchar('2');} else if(number[j]==3) {putchar('3');} else if(number[j]==4) {putchar('4');} else if(number[j]==5) {putchar('5');} else if(number[j]==6) {putchar('6');} else if(number[j]==7) {putchar('7');} else if(number[j]==8) {putchar('8');} else if(number[j]==9) {putchar('9');} else if(number[j]==10) {putchar('A');} else if(number[j]==11) {putchar('B');} else if(number[j]==12) {putchar('C');} else if(number[j]==13) {putchar('D');} else if(number[j]==14) {putchar('E');} else {putchar('F');} } putchar('\n'); }