title: exited with non-zero status
body: <p>I am new in programming and try to learn by own ,i have an error in my code,i don't understand it's a syntax error or i've done smth wrong.I used 3 equations with it's condition and made it in for isntruction,while,do-while,if-else,with switch.After i introduce variable a it show me  an error "exited with non-zero status".</p>
  
  <pre><code>#include &lt;stdio.h&gt;
  #include &lt;math.h&gt;
  
  int main() {
      float a,x,b;
      float L;
      int m;
  
      printf("Enter variables a,x,b:");
      scanf("%f%f%f,&amp;a,&amp;x,&amp;b");
      printf("For using for instruction (enter 1)");
      printf("For using while instruction (enter 2)");
      printf("For using do-while instruction (enter 3)");
      printf("For using ef instruction (enter 4)");
      scanf("%d,&amp;m");
      switch(m){
          case 1:
              for (x=0;x&lt;1.2;x++)
              {
                  L=2*cos(x-(3.14/6));
              }
              for (x=0;x&gt;= 1.2 &amp;&amp; x&lt;=3.9;x++)
              {
                  L=x*x/(a+cos(powf((x+b),3)));
              }
              for (x=0;x&gt;3.9;x++)
              {
                  L=fabs(x/2*a)+powf(sin(x+1),2);
              }
              break;
          case 2:
              while (x&lt;1.2)
              {
                   L=2*cos(x-(3.14/6));
              }
              while (x&gt;= 1.2 &amp;&amp; x&lt;=3.9)
              {
                   L=x*x/(a+cos(powf((x+b),3)));
              }
              while (x&gt;3.9)
              {
                   L=fabs(x/2*a)+powf(sin(x+1),2);
              }
              break;
         case 3:
              do 
              {
                  L=2*cos(x-(3.14/6));
              }
              while (x&lt;1.2);
              do 
              {
                  L=x*x/(a+cos(powf((x+b),3)));
              }
              while (x&gt;= 1.2 &amp;&amp; x&lt;=3.9);
              do
              {
                  L=fabs(x/2*a)+powf(sin(x+1),2);
              }
              while (x&gt;3.9);
              break;
         case 4:
             if (x&lt;1.2)
             {
                 L=2*cos(x-(3.14/6));
             }
             else
             {
                 printf("First statement is false");
             }
             if(x&gt;= 1.2 &amp;&amp; x&lt;=3.9)
             {
                 L=x*x/(a+cos(powf((x+b),3)));
             }
             else
             {
                 printf("Second statement is false");
             }
             if(x&gt;3.9)
             {
                 L=fabs(x/2*a)+powf(sin(x+1),2);
             }
             else
             {
                 printf("Third statement is false");
             }
             break;
         default:
             printf("\nNo right choices\n");
     }
     printf("Your answer is: L = %.3f,L");
  }
  </code></pre>