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https://github.com/byteworksinc/ORCA-C.git
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3c2b492618
The basic approach is to generate a single expression tree containing the code for the initialization plus the reference to the compound literal (or its address). The various subexpressions are joined together with pc_bno pcodes, similar to the code generated for the comma operator. The initializer expressions are placed in a balanced binary tree, so that it is not excessively deep. Note: Common subexpression elimination has poor performance for very large trees. This is not specific to compound literals, but compound literals for relatively large arrays can run into this issue. It will eventually complete and generate a correct program, but it may be quite slow. To avoid this, turn off CSE.
66 lines
1.5 KiB
C
66 lines
1.5 KiB
C
/*
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* Test of compound literals (C99).
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*/
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#include <stdio.h>
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int *p = (int[]){1,2,3};
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int *q = &(int[100]){4,5,6}[1];
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struct S *s = &(struct S {int i; double d; void *p;}){100,200.5,&p};
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int f(struct S s) {
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return s.i;
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}
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double g(struct S *s) {
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return s->d + s->i;
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}
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int main(void) {
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if (p[2] != 3)
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goto Fail;
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if (*q != 5)
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goto Fail;
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if (q[80] != 0)
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goto Fail;
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p[2] = s->i;
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if (p[2] != 100)
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goto Fail;
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if ((int[]){6,7,8}[2] != 8)
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goto Fail;
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if (((char){34} += (long long){53}) != 87)
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goto Fail;
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if ((int){(double){(long){(char){22}}}} != (signed char){22})
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goto Fail;
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if (((struct S*)((struct S){0,-.5,&(struct S){-12,14,0}}.p))->d != 14.)
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goto Fail;
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if (f((struct S){f((struct S){-12,14,0}),23.5}) != -12)
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goto Fail;
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if (g(&(struct S){5,2.5,&(char){7}}) != 7.5)
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goto Fail;
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if ((char[100]){12}[99] != 0)
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goto Fail;
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if ((char[]){"Hello world"}[10] != 'd')
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goto Fail;
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if ((char[100]){"Hello world"}[50] != '\0')
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goto Fail;
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printf ("Passed Conformance Test c99complit\n");
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return 0;
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Fail:
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printf ("Failed Conformance Test c99complit\n");
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}
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