glsl: remove unused state variable

Because compile_shader function state variable
not determine whether the compilation is successful.

Signed-off-by: cheyang <cheyang@bytedance.com>
Reviewed-by: Tapani Pälli <tapani.palli@intel.com>
Part-of: <https://gitlab.freedesktop.org/mesa/mesa/-/merge_requests/8178>

diff --git a/src/compiler/glsl/standalone.cpp b/src/compiler/glsl/standalone.cpp
index 653f3801e2bf45d98faacbc6587df9fbb654d331..b34583e54bd6b9e45f5a103502d4073f8fe7bf8b 100644 (file)
--- a/src/compiler/glsl/standalone.cpp
+++ b/src/compiler/glsl/standalone.cpp
@@ -380,15 +380,12 @@ load_text_file(void *ctx, const char *file_name)
 static void
 compile_shader(struct gl_context *ctx, struct gl_shader *shader)
 {
-   struct _mesa_glsl_parse_state *state =
-      new(shader) _mesa_glsl_parse_state(ctx, shader->Stage, shader);
-
    _mesa_glsl_compile_shader(ctx, shader, options->dump_ast,
                              options->dump_hir, true);
 
    /* Print out the resulting IR */
-   if (!state->error && options->dump_lir) {
-      _mesa_print_ir(stdout, shader->ir, state);
+   if (shader->CompileStatus == COMPILE_SUCCESS && options->dump_lir) {
+      _mesa_print_ir(stdout, shader->ir, NULL);
    }
 
    return;