Here is my program :
#include %26lt;stdio.h%26gt;
#include %26lt;math.h%26gt;
void drain();
int main()
{
double x, y, Y, RE;
printf("\nEnter x: ");
scanf("%lf", %26amp;x);
y = pow(exp(x), 1.5);
printf("True value of exp(1.5*%.4f) = %lf", x, y);
char ch;
fflush(stdin);
printf("\n\nEnter a character 1-4 (0 to exit):\n");
ch = getchar();
switch(ch){
case 1:
printf("1 term(s) approximation"};
Y = pow(x, 0.)/1.;
printf("Approximate exp(1.5*%.4f) = ", x);
RE = 100 * (Y - y)/y;
printf("\nRelative error = %lf", RE);
break;
default:
printf("unrecognized operator");
}
}
.c: In function `main':
:17: error: parse error before '}' token
What should i do? .... thx so much
C++ I have an error that I couldn't figure out T_T?
well, thts easy, u see, u have defined main as returning an integer but u r not returning anything
check out the same program wid the corrections below
#include %26lt;stdio.h%26gt;
#include %26lt;math.h%26gt;
void drain();
int main()
{
double x, y, Y, RE;
printf("\nEnter x: ");
scanf("%lf", %26amp;x);
y = pow(exp(x), 1.5);
printf("True value of exp(1.5*%.4f) = %lf", x, y);
char ch;
fflush(stdin);
printf("\n\nEnter a character 1-4 (0 to exit):\n");
ch = getchar();
switch(ch){
case 1:
printf("1 term(s) approximation"};
Y = pow(x, 0.)/1.;
printf("Approximate exp(1.5*%.4f) = ", x);
RE = 100 * (Y - y)/y;
printf("\nRelative error = %lf", RE);
break;
default:
printf("unrecognized operator");
}
return(0);
}
if my solution was of help, gimme the best answer vote, pls
Reply:My vote goes to this line!
printf("1 term(s) approximation"};
It is line 17!
The end character should be ) NOT }
That was fun.
Reply:switch(ch) { %26lt;!-- put a space between (ch) and {
case 1:
printf("1 term(s) approximation"};
Y = pow(x, 0.)/1.; %26lt;!-- why do u have a point (.) after 1? if its a decimal put 1.0 and why do u havea period after 0?
printf("Approximate exp(1.5*%.4f) = ", x);
RE = 100 * (Y - y)/y;
printf("\nRelative error = %lf", RE);
break;
default:
printf("unrecognized operator");
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