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RENESAS TOOL NEWS [RSO-M3T-CC32R-030601D]
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Notes on Using Cross-Tool Kit M3T-CC32R
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Please take note of the following problems in using the M3T-CC32R cross-tool kit for the M32R family MCUs:
- On referencing and updating variables in an infinite loop
- On converting a value by a printf-like standard library function
- Problem on Referencing and Updating Variables in an Infinite Loop
- 1.1 Version Concerned
- M3T-CC32R V.4.00 Release 1
- 1.2 Description
- If a variable is referenced and then updated in an infinite loop, the previously referenced value of the variable may not be updated.
- 1.3 Conditions
- This problem occurs if the following six conditions are satisfied:
| (1) | Any optimizing option covering the -O5 or -O6 option's function (-O5, -O6, -O7, -Otime only or -Ospace only) is selected at compilation.
| Note: | The -O7, -Otime only and -Ospace only options include the functions of the -O5 and -O6 options. |
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| (2) | Either of the following loop constructs exists in the program:
| (a) | a while, do-while, or for statement whose continuation condition is always TRUE |
| (b) | a loop made by an unconditionally executed goto statement |
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| (3) | In the loop in (2) exists no break, goto or return statement used to exit from the loop. |
| (4) | A variable is referenced in the loop in (2). |
| (5) | The variable in (4) is updated after referenced in the loop in (2). |
| (6) | This variable is not referenced after the update in (5) until the loop is ended. |
- 1.4 Examples
Source file 1:
-----------------------------------------------------------------
void func1(int);
void
foo1(void)
{
int a = 0;
while (1) { /* Conditions(2)(a) and (3) */
func1(a); /* Condition (4) */
++a; /* Condition (5) */
/* Condition (6) */
}
}
-----------------------------------------------------------------
Source file 2:
-----------------------------------------------------------------
int func2(int);
void
foo2(void)
{
int a = 0;
for (;;) { /* Conditions (2)(a) and (3) */
a = func2(a) + 1; /* Conditions (4) and (5) */
/* Condition (6) */
}
}
-----------------------------------------------------------------
Source file 3:
-----------------------------------------------------------------
volatile int x;
void
foo3(void)
{
int a = 0;
infinite_loop: /* Conditions (2)(b) and (3) */
x = a; /* Condition (4) */
++a; /* Condition (5) */
goto infinite_loop; /* Conditions (2)(b) and (3) */
/* Condition (6) */
}
-----------------------------------------------------------------
- 1.5 Solution
- This problem has been fixed in the M3T-CC32R V.4.10 Release 1. So please download the fixed product from HERE.
- 1.6 Workaround in Using the M3T-CC32R V.4.00 Release 1
- If you continue to use the M3T-CC32R V.4.00 Release 1, this problem can be circumvented in either of the following ways:
| (1) | Suppress the optimization equivalent to the -O5 or -O6 option. |
| (2) | As the continuation condition, use an allocated address.
That is, after defining a variable, allocate a fixed address to it using the #pragma ADDRESS directive; then reference the address allocated to the variable (the allocated address) to use it as the continuation condition. |
Modification of Source file 1:
-----------------------------------------------------------------
void func1(int);
int dummy1;
#pragma ADDRESS dummy1 1 /* Allocate address 1 to dummy1 */
void
foo1(void)
{
int a = 0;
while (&dummy1) { /*Use &dummy1 as continuation condition*/
func1(a);
++a;
}
}
-----------------------------------------------------------------
Modification of Source file 2:
-----------------------------------------------------------------
int func2(int);
int dummy2;
#pragma ADDRESS dummy2 1 /* Allocate address 1 to dummy2 */
void
foo2(void)
{
int a = 0;
for (;&dummy2;) { /*Use &dummy2 as continuation condition*/
a = func2(a) + 1;
}
}
-----------------------------------------------------------------
Modification of Source file 3:
-----------------------------------------------------------------
volatile int x;
int dummy3;
#pragma ADDRESS dummy3 1 /* Allocate address 1 to dummy3 */
void
foo3(void)
{
int a = 0;
infinite_loop:
x = a;
++a;
if (&dummy3) /* Use &dummy3 as continuation condition */
goto infinite_loop;
}
-----------------------------------------------------------------
Notes on Using Cross-Tool Kit M3T-CC32R
RENESAS TOOL NEWS [RSO-M3T-CC32R-030601D]
- Problem on converting a value by a printf-like standard library function
- 2.1 Versions Concerned
- M3T-CC32R V.1.00 Release 1 -- V.4.10 Release 1
- 2.2 Description
- When a value less than the value which can be calculated from the specified precision is converted by the %f specification of a format output function such as printf, usually the number of zeros satisfying the precision given by the %f specification is outputted in decimal places. However, the number of zeros may be less by one than the satisfactory number of them depending on the value to be converted.
- 2.3 Conditions
- This problem occurs if the following two conditions are satisfied:
| (1) | The %f specification of a printf-like function (printf, vprintf, fprintf, vfprintf, sprintf, or vsprintf) is used to convert a value. |
| (2) | When the precision given by the %f specification in (1) is n, the value to be converted by a printf-like function is equal to or less than ten to the power of negative n.
Note that if no precision is explicitly specified in %f, six is used as the precision. |
- 2.4 Examples
Source file:
-----------------------------------------------------------------
#include <stdio.h>
char bf1[32], bf2[32], bf3[32], bf4[32], bf5[32], bf6[32];
int main()
{
double x;
x = 0.01; /* Condition (2) */
sprintf(bf1, "%5.1f\n", x); /* Conditions (1) and (2)
with precision 1 */
/* A program containing bf1 */
x = 0.005; /* Condition (2) */
sprintf(bf2, "%5.1f\n", x); /* Conditions (1) and (2)
with precision 1 */
/* A program containing bf2 */
x = 0.0064; /* Condition (2) */
sprintf(bf3, "%5.1f\n", x); /* Conditions (1) and (2)
with precision 1 */
/* A program containing bf3 */
x = 0.0007; /* Condition (2) */
sprintf(bf4, "%5.2f\n", x); /* Conditions (1) and (2)
with precision 2 */
/* A program containing bf4 */
x = 8.1e-5; /* Condition (2) */
sprintf(bf5, "%6.2f\n", x); /* Conditions (1) and (2)
with precision 2 */
/* A program containing bf5 */
x = 7.3e-8; /* Condition (2) */
sprintf(bf6, "%10f\n", x); /* Conditions (1) and (2)
with precision 6 */
/* A program containing bf6 */
return 0;
}
-----------------------------------------------------------------
- The output results of the above conversions (the contents of arrays bf1 through bf6)
Output Result Correct Output
bf1: " 0." " 0.0"
bf2: " 0." " 0.0"
bf3: " 0." " 0.0"
bf4: " 0.0" " 0.00"
bf5: " 0.0" " 0.00"
bf6: " 0.00000" " 0.000000"
- 2.5 Workaround
- Discard the decimal places exceeding the specified precision using the floor standard library function in the math.h header file.
- Example:
-
| Modify | printf("%5.1\n, x) |
| to | printf("%5.1\n, floor(x * 1E1 + 0.5)/1E1) |
- Note:
- The portion of 1E1 varies according to precisions.
Change %5.2f (precision 2) to floor(x * 1E2 + 0.5)/1E2
Change %8f (precision 6) to floor(x * 1E6 + 0.5)/1E6
Modification of Source file 1:
-----------------------------------------------------------------
#include <stdio.h>
#include <math.h> /* Declaration for using the floor function */
char bf1[32], bf2[32], bf3[32], bf4[32], bf5[32], bf6[32];
int main()
{
double x;
x = 0.01;
/* The decimal places exceeding the specified precision
discarded in variable x (the same in the later) */
sprintf(bf1, "%5.1f\n", floor(x * 1E1 + 0.5)/1E1);
/* A program containing bf1 */
x = 0.005;
sprintf(bf2, "%5.1f\n", floor(x * 1E1 + 0.5)/1E1);
/* A program containing bf2 */
x = 0.0064;
sprintf(bf3, "%5.1f\n", floor(x * 1E1 + 0.5)/1E1);
/* A program containing bf3 */
x = 0.0007;
sprintf(bf4, "%5.2f\n", floor(x * 1E2 + 0.5)/1E2);
/* A program containing bf4 */
x = 8.1e-5;
sprintf(bf5, "%6.2f\n", floor(x * 1E2 + 0.5)/1E2);
/* A program containing bf5 */
x = 7.3e-8;
sprintf(bf6, "%10f\n", floor(x * 1E6 + 0.5)/1E6);
/* A program containing bf6 */
return 0;
}
-----------------------------------------------------------------
The output results of the above conversions
Output Result
bf1: " 0.0"
bf2: " 0.0"
bf3: " 0.0"
bf4: " 0.00"
bf5: " 0.00"
bf6: " 0.000000"
- 2.6 Schedule of Fixing the Problem
- We plan to fix this problem in our next release of the product.
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