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The following noncompliant code demonstrates the perils of operating on data of incompatible types. An attempt is made to increment an integer type cast to a floating point type and a floating point cast to an integer type.
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float f = 0.0;
int i = 0;
float *fp;
int *ip;
assert(sizeof(int) == sizeof(float));
ip = (int*) &f;
fp = (float*) &i;
printf("int is %d, float is %f\n", i, f);
(*ip)++;
(*fp)++;
printf("int is %d, float is %f\n", i, f);
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In this compliant solution, the pointers are assigned to reference variables of compatible data types.
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float f = 0.0;
int i = 0;
float *fp;
int *ip;
ip = &i;
fp = &f;
printf("int is %d, float is %f\n", i, f);
(*ip)++;
(*fp)++;
printf("int is %d, float is %f\n", i, f);
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The following code behaves differently depending on whether the implementation is left-to-right or right-to-left:
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struct bf {
unsigned int m1 : 8;
unsigned int m2 : 8;
unsigned int m3 : 8;
unsigned int m4 : 8;
}; /* 32 bits total */
void function() {
struct bf data;
unsigned char *ptr;
data.m1 = 0;
data.m2 = 0;
data.m3 = 0;
data.m4 = 0;
ptr = (unsigned char *)&data;
(*ptr)++; /* can increment data.m1 or data.m4 */
}
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This compliant solution is explicit in which fields it modifies.
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struct bf {
unsigned int m1 : 8;
unsigned int m2 : 8;
unsigned int m3 : 8;
unsigned int m4 : 8;
}; /* 32 bits total */
void function() {
struct bf data;
data.m1 = 0;
data.m2 = 0;
data.m3 = 0;
data.m4 = 0;
data.m1++;
}
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In the following noncompliant code, assuming eight bits to a byte, if bit-fields of six and four bits are declared, is each bit-field contained within a byte, or are the bit-fields split across multiple bytes?
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struct bf {
unsigned int m1 : 6;
unsigned int m2 : 4;
};
void function() {
unsigned char *ptr;
struct bf data;
data.m1 = 0;
data.m2 = 0;
ptr = (unsigned char *)&data;
ptr++;
*ptr += 1; /* what does this increment? */
}
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This compliant solution is explicit in which fields it modifies.
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struct bf {
unsigned int m1 : 6;
unsigned int m2 : 4;
};
void function() {
struct bf data;
data.m1 = 0;
data.m2 = 0;
data.m2 += 1;
}
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